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proofwiki-6900
Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication
:$p \implies \neg q \vdash \neg \paren {p \land q}$
{{BeginTableau|p \implies \neg q \vdash \neg \paren {p \land q} }} {{Premise|1|p \implies \neg q}} {{Assumption|2|p \land q|Assume the opposite of what is to be proved ...}} {{Simplification|3|2|p|2|1}} {{Simplification|4|2|q|2|2}} {{ModusPonens|5|1, 2|\neg q|1|3}} {{NonContradiction|6|1, 2|4|5|... and demonstrate a co...
:$p \implies \neg q \vdash \neg \paren {p \land q}$
{{BeginTableau|p \implies \neg q \vdash \neg \paren {p \land q} }} {{Premise|1|p \implies \neg q}} {{Assumption|2|p \land q|Assume the opposite of what is to be proved ...}} {{Simplification|3|2|p|2|1}} {{Simplification|4|2|q|2|2}} {{ModusPonens|5|1, 2|\neg q|1|3}} {{NonContradiction|6|1, 2|4|5|... and demonstrate a co...
Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication
https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1/Reverse_Implication
[ "Modus Ponendo Tollens" ]
[]
[ "Category:Modus Ponendo Tollens" ]
proofwiki-6901
Modus Ponendo Tollens/Variant/Formulation 1
: $\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||cccc|} \hline \neg & (p & \land & q) & p & \implies & \neg & q \\ \hline T & F & F & F & F & T & T & F \\ T & F & F & T & F & T & F & T \\ T & T & F & F & T & T & T & F \\ F & T & T & T...
: $\neg \left({p \land q}\right) \dashv \vdash p \implies \neg q$
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|cccc||cccc|} \hline \neg & (p & \land & q) & p & \implies & \neg & q \\ ...
Modus Ponendo Tollens/Variant/Formulation 1/Proof
https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1
https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_1/Proof
[ "Modus Ponendo Tollens" ]
[]
[ "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6902
Modus Ponendo Tollens/Variant/Formulation 2
:$\vdash \paren {\neg \paren {p \land q} } \iff \paren {p \implies \neg q}$
{{BeginTableau|\vdash \paren {\neg \paren {p \land q} } \iff \paren {p \implies \neg q} }} {{Assumption|1|\neg \paren {p \land q} }} {{SequentIntro|2|1|p \implies \neg q|1|Modus Ponendo Tollens: Formulation 1: Forward Implication}} {{Implication|3||\paren {\neg \paren {p \land q} } \implies \paren {p \implies \neg q}|1...
:$\vdash \paren {\neg \paren {p \land q} } \iff \paren {p \implies \neg q}$
{{BeginTableau|\vdash \paren {\neg \paren {p \land q} } \iff \paren {p \implies \neg q} }} {{Assumption|1|\neg \paren {p \land q} }} {{SequentIntro|2|1|p \implies \neg q|1|[[Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication|Modus Ponendo Tollens: Formulation 1: Forward Implication]]}} {{Implication|3||\pa...
Modus Ponendo Tollens/Variant/Formulation 2
https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_2
https://proofwiki.org/wiki/Modus_Ponendo_Tollens/Variant/Formulation_2
[ "Modus Ponendo Tollens" ]
[]
[ "Modus Ponendo Tollens/Variant/Formulation 1/Forward Implication", "Modus Ponendo Tollens/Variant/Formulation 1/Reverse Implication" ]
proofwiki-6903
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication
:$\neg \paren {p \land \neg q} \vdash p \implies q$
{{BeginTableau|\neg \paren {p \land \neg q} \vdash p \implies q}} {{Premise|1|\neg \paren {p \land \neg q} }} {{DeMorgan|2|1|\neg p \lor \neg \neg q|1|Disjunction of Negations}} {{SequentIntro|3|1|p \implies \neg \neg q|2|Rule of Material Implication}} {{Assumption|4|p}} {{ModusPonens|5|1, 4|\neg \neg q|3|4}} {{DoubleN...
:$\neg \paren {p \land \neg q} \vdash p \implies q$
{{BeginTableau|\neg \paren {p \land \neg q} \vdash p \implies q}} {{Premise|1|\neg \paren {p \land \neg q} }} {{DeMorgan|2|1|\neg p \lor \neg \neg q|1|Disjunction of Negations}} {{SequentIntro|3|1|p \implies \neg \neg q|2|[[Rule of Material Implication]]}} {{Assumption|4|p}} {{ModusPonens|5|1, 4|\neg \neg q|3|4}} {{Dou...
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Reverse Implication
https://proofwiki.org/wiki/Conditional_is_Equivalent_to_Negation_of_Conjunction_with_Negative/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/Conditional_is_Equivalent_to_Negation_of_Conjunction_with_Negative/Formulation_1/Reverse_Implication
[ "Conditional is Equivalent to Negation of Conjunction with Negative" ]
[]
[ "Rule of Material Implication" ]
proofwiki-6904
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1
:$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccccc|} \hline p & \implies & q & \neg & (p & \land & \neg & q) \\ \hline \F & \T & \F & \T & \F & \F & \T & \F \\ \F & \T & \T & \T & \F & \F & \F &...
:$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccc||ccccc|} \hline p & \impli...
Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Conditional_is_Equivalent_to_Negation_of_Conjunction_with_Negative/Formulation_1
https://proofwiki.org/wiki/Conditional_is_Equivalent_to_Negation_of_Conjunction_with_Negative/Formulation_1/Proof_by_Truth_Table
[ "Conditional is Equivalent to Negation of Conjunction with Negative" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6905
Zero Derivative implies Constant Complex Function
Let $D \subseteq \C$ be a connected domain of $\C$. Let $f: D \to \C$ be a complex-differentiable function. For all $z \in D$, let $\map {f'} z = 0$. Then $f$ is constant on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two real-valued functions defined in the Cauchy-Riemann Equations: :$\map u {x, y} = \map \Re {\map f z}$ :$\map v {x, y} = \map \Im {\map f z}$ By the Cauchy-Riemann Equations: :$f' = \dfrac {\partial f} {\partial x} = \dfrac {\partial u} {\par...
Let $D \subseteq \C$ be a [[Definition:Connected Domain (Complex Analysis)|connected domain]] of $\C$. Let $f: D \to \C$ be a [[Definition:Complex-Differentiable Function|complex-differentiable function]]. For all $z \in D$, let $\map {f'} z = 0$. Then $f$ is [[Definition:Constant Mapping|constant]] on $D$.
Let $u, v: \set {\tuple {x, y} \in \R^2: x + i y = z \in D} \to \R$ be the two [[Definition:Real-Valued Function|real-valued functions]] defined in the [[Cauchy-Riemann Equations]]: :$\map u {x, y} = \map \Re {\map f z}$ :$\map v {x, y} = \map \Im {\map f z}$ By the [[Cauchy-Riemann Equations]]: :$f' = \dfrac {\pa...
Zero Derivative implies Constant Complex Function
https://proofwiki.org/wiki/Zero_Derivative_implies_Constant_Complex_Function
https://proofwiki.org/wiki/Zero_Derivative_implies_Constant_Complex_Function
[ "Complex Analysis" ]
[ "Definition:Connected Domain (Complex Analysis)", "Definition:Differentiable Mapping/Complex Function", "Definition:Constant Mapping" ]
[ "Definition:Real-Valued Function", "Cauchy-Riemann Equations", "Cauchy-Riemann Equations", "Zero Derivative implies Constant Function", "Definition:Partial Derivative", "Connected Domain is Connected by Staircase Contours", "Definition:Staircase Contour", "Definition:Contour/Endpoints/Complex Plane", ...
proofwiki-6906
Relation between P-Product Metric and Chebyshev Distance on Real Vector Space
For $n \in \N$, let $\R^n$ be a Euclidean space. Let $p \in \R_{\ge 1}$. Let $d_p$ be the $p$-product metric on $\R^n$. Let $d_\infty$ be the Chebyshev distance on $\R^n$. Then :$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_p} {x, y} \le n^{1/p} \map {d_\infty} {x, y}$
By definition of the Chebyshev distance on $\R^n$, we have: :$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$ where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$. Let $j$ be chosen so that: :$\ds \size {x_j - y_j} = \max_{i \mathop = 1}^n \size {x_i - y_i}$ Then: {...
For $n \in \N$, let $\R^n$ be a [[Definition:Euclidean Space|Euclidean space]]. Let $p \in \R_{\ge 1}$. Let $d_p$ be the [[Definition:P-Product Metric on Real Vector Space|$p$-product metric]] on $\R^n$. Let $d_\infty$ be the [[Definition:Chebyshev Distance on Real Vector Space|Chebyshev distance]] on $\R^n$. The...
By definition of the [[Definition:Chebyshev Distance on Real Vector Space|Chebyshev distance]] on $\R^n$, we have: :$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$ where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$. Let $j$ be chosen so that: :$\ds \size {x_j - ...
Relation between P-Product Metric and Chebyshev Distance on Real Vector Space
https://proofwiki.org/wiki/Relation_between_P-Product_Metric_and_Chebyshev_Distance_on_Real_Vector_Space
https://proofwiki.org/wiki/Relation_between_P-Product_Metric_and_Chebyshev_Distance_on_Real_Vector_Space
[ "Chebyshev Distance", "P-Product Metrics on Real Vector Space are Topologically Equivalent" ]
[ "Definition:Euclidean Space", "Definition:P-Product Metric/Real Vector Space", "Definition:Chebyshev Distance/Real Vector Space" ]
[ "Definition:Chebyshev Distance/Real Vector Space", "Power of Maximum is not Greater than Sum of Powers", "Sum of r Powers is not Greater than r times Power of Maximum" ]
proofwiki-6907
P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case
For $n \in \N$, let $\R^n$ be a real vector Space. Let $r, t \in \R_{\ge 1}$. Let $d_r$ and $d_t$ be $p$-product metrics on $\R^n$. Then $d_r$ and $d_t$ are topologically equivalent.
We show that $\ds \map {d_r} {x, y} \ge \map {d_t} {x, y}$, which is equivalent to proving that: :$\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r} \ge \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^t}^{1/t}$ Let $\forall i \in \closedint 1 n: s_i = \size {x_i - y_i}$. Suppose $s_k = 0$ for some $k \in \...
For $n \in \N$, let $\R^n$ be a [[Definition:Real Vector Space|real vector Space]]. Let $r, t \in \R_{\ge 1}$. Let $d_r$ and $d_t$ be [[Definition:P-Product Metric on Real Vector Space|$p$-product metrics]] on $\R^n$. Then $d_r$ and $d_t$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent]].
We show that $\ds \map {d_r} {x, y} \ge \map {d_t} {x, y}$, which is equivalent to proving that: :$\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r} \ge \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^t}^{1/t}$ Let $\forall i \in \closedint 1 n: s_i = \size {x_i - y_i}$. Suppose $s_k = 0$ for some $k \...
P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case
https://proofwiki.org/wiki/P-Product_Metrics_on_Real_Vector_Space_are_Topologically_Equivalent/Inequality_for_General_Case
https://proofwiki.org/wiki/P-Product_Metrics_on_Real_Vector_Space_are_Topologically_Equivalent/Inequality_for_General_Case
[ "P-Product Metrics on Real Vector Space are Topologically Equivalent" ]
[ "Definition:Real Vector Space", "Definition:P-Product Metric/Real Vector Space", "Definition:Topologically Equivalent Metrics" ]
[ "Derivative of Function to Power of Function", "Derivative of Exponential Function", "Sum Rule for Derivatives", "Power Rule for Derivatives", "Logarithm of Power", "Natural Logarithm of 1 is 0", "Logarithm is Strictly Increasing", "Derivative of Monotone Function", "Definition:Decreasing/Real Funct...
proofwiki-6908
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication
:$\neg p \lor \neg q \vdash \neg \paren {p \land q}$
{{BeginTableau|\neg p \lor \neg q \vdash \neg \paren {p \land q} }} {{Premise|1|\neg p \lor \neg q}} {{Assumption|2|p \land q}} {{Simplification|3|2|p|2|1}} {{Simplification|4|2|q|2|2}} {{Assumption|5|\neg p}} {{NonContradiction|6|2, 5|3|5}} {{Assumption|7|\neg q}} {{NonContradiction|8|2, 7|4|7}} {{ProofByCases|9|1, 2|...
:$\neg p \lor \neg q \vdash \neg \paren {p \land q}$
{{BeginTableau|\neg p \lor \neg q \vdash \neg \paren {p \land q} }} {{Premise|1|\neg p \lor \neg q}} {{Assumption|2|p \land q}} {{Simplification|3|2|p|2|1}} {{Simplification|4|2|q|2|2}} {{Assumption|5|\neg p}} {{NonContradiction|6|2, 5|3|5}} {{Assumption|7|\neg q}} {{NonContradiction|8|2, 7|4|7}} {{ProofByCases|9|1, 2|...
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1/Forward_Implication
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6909
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication
:$\neg \paren {p \land q} \vdash \neg p \lor \neg q$
{{BeginTableau|\neg \paren {p \land q} \vdash \neg p \lor \neg q}} {{Premise|1|\neg \paren {p \land q} }} {{Assumption|2|\neg \paren {\neg p \lor \neg q} }} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor \neg q|3|1}} {{NonContradiction|5|2, 3|4|2}} {{Reductio|6|2|p|3|5}} {{Assumption|7|\neg q}} {{Addition|8|7|\neg ...
:$\neg \paren {p \land q} \vdash \neg p \lor \neg q$
{{BeginTableau|\neg \paren {p \land q} \vdash \neg p \lor \neg q}} {{Premise|1|\neg \paren {p \land q} }} {{Assumption|2|\neg \paren {\neg p \lor \neg q} }} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor \neg q|3|1}} {{NonContradiction|5|2, 3|4|2}} {{Reductio|6|2|p|3|5}} {{Assumption|7|\neg q}} {{Addition|8|7|\neg ...
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication/Proof 1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1/Reverse_Implication/Proof_1
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6910
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1
:$\neg p \lor \neg q \dashv \vdash \neg \paren {p \land q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :<nowiki>$\begin{array}{|ccccc||cccc|} \hline \neg & p & \lor & \neg & q & \neg & (p & \land & q) \\ \hline \T & \F & \T & \T & \F & \T & \F & \F & \F \\ \T & \F & \T & \...
:$\neg p \lor \neg q \dashv \vdash \neg \paren {p \land q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|ccccc||cccc|} \hline ...
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6911
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1
:$\neg p \lor \neg q \dashv \vdash \neg \paren {p \land q}$
{{BeginTableau|\neg \paren {p \land q} \vdash \neg p \lor \neg q}} {{Premise|1|\neg \paren {p \land q} }} {{Assumption|2|\neg \paren {\neg p \lor \neg q} }} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor \neg q|3|1}} {{NonContradiction|5|2, 3|4|2}} {{Reductio|6|2|p|3|5}} {{Assumption|7|\neg q}} {{Addition|8|7|\neg ...
:$\neg p \lor \neg q \dashv \vdash \neg \paren {p \land q}$
{{BeginTableau|\neg \paren {p \land q} \vdash \neg p \lor \neg q}} {{Premise|1|\neg \paren {p \land q} }} {{Assumption|2|\neg \paren {\neg p \lor \neg q} }} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor \neg q|3|1}} {{NonContradiction|5|2, 3|4|2}} {{Reductio|6|2|p|3|5}} {{Assumption|7|\neg q}} {{Addition|8|7|\neg ...
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication/Proof 1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_1/Reverse_Implication/Proof_1
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6912
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2
:$\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} }$
{{BeginTableau|\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} } }} {{Assumption|1|\neg p \lor \neg q}} {{SequentIntro|2|1|\neg \paren {p \land q}|1|De Morgan's Laws (Logic): Disjunction of Negations: Formulation 1}} {{Implication|3||\paren {\neg p \lor \neg q} \implies \paren {\neg \paren {p \l...
:$\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} }$
{{BeginTableau|\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} } }} {{Assumption|1|\neg p \lor \neg q}} {{SequentIntro|2|1|\neg \paren {p \land q}|1|[[De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication|De Morgan's Laws (Logic): Disjunction of Negations: Formulati...
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2/Proof 1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_2/Proof_1
[ "De Morgan's Laws (Logic)" ]
[]
[ "De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication", "De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication" ]
proofwiki-6913
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2
:$\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} }$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. :<nowiki>$\begin{array}{|ccccc|c|cccc|} \hline \neg & p & \lor & \neg & q & \iff & \neg & (p & \land & q) \\ \hline \T & \F & \T & \T & \F & \T & \T & \F & \F & \F \\ \T...
:$\vdash \paren {\neg p \lor \neg q} \iff \paren {\neg \paren {p \land q} }$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|...
De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction_of_Negations/Formulation_2/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6914
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication
:$\neg p \land \neg q \vdash \neg \paren {p \lor q}$
{{BeginTableau|\neg p \land \neg q \vdash \neg \paren {p \lor q} }} {{Premise|1|\neg p \land \neg q}} {{Simplification|2|1|\neg p|1|1}} {{Simplification|3|1|\neg q|1|2}} {{Assumption|4|p \lor q}} {{Assumption|5|p}} {{NonContradiction|6|1, 5|5|2}} {{Assumption|7|q}} {{NonContradiction|8|1, 7|7|3}} {{ProofByCases|9|1, 4|...
:$\neg p \land \neg q \vdash \neg \paren {p \lor q}$
{{BeginTableau|\neg p \land \neg q \vdash \neg \paren {p \lor q} }} {{Premise|1|\neg p \land \neg q}} {{Simplification|2|1|\neg p|1|1}} {{Simplification|3|1|\neg q|1|2}} {{Assumption|4|p \lor q}} {{Assumption|5|p}} {{NonContradiction|6|1, 5|5|2}} {{Assumption|7|q}} {{NonContradiction|8|1, 7|7|3}} {{ProofByCases|9|1, 4|...
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_1/Forward_Implication
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6915
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication
:$\neg \paren {p \lor q} \vdash \neg p \land \neg q$
{{BeginTableau|\neg \paren {p \lor q} \vdash \neg p \land \neg q}} {{Premise|1|\neg \paren {p \lor q} }} {{Assumption|5|p}} {{Addition|3|2|p \lor q|2|1}} {{NonContradiction|4|1, 3|3|1}} {{Contradiction|5|1|\neg p|2|4}} {{Assumption|6|q}} {{Addition|7|6|p \lor q|6|2}} {{NonContradiction|8|1, 7|7|1}} {{Contradiction|9|1|...
:$\neg \paren {p \lor q} \vdash \neg p \land \neg q$
{{BeginTableau|\neg \paren {p \lor q} \vdash \neg p \land \neg q}} {{Premise|1|\neg \paren {p \lor q} }} {{Assumption|5|p}} {{Addition|3|2|p \lor q|2|1}} {{NonContradiction|4|1, 3|3|1}} {{Contradiction|5|1|\neg p|2|4}} {{Assumption|6|q}} {{Addition|7|6|p \lor q|6|2}} {{NonContradiction|8|1, 7|7|1}} {{Contradiction|9|1|...
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_1/Reverse_Implication
[ "De Morgan's Laws (Logic)" ]
[]
[ "Category:De Morgan's Laws (Logic)" ]
proofwiki-6916
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1
:$\neg p \land \neg q \dashv \vdash \neg \paren {p \lor q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :<nowiki>$\begin{array}{|ccccc||cccc|} \hline \neg & p & \land & \neg & q & \neg & (p & \lor & q) \\ \hline \T & \F & \T & \T & \F & \T & \F & \F & \F \\ \T & \F & \F & \...
:$\neg p \land \neg q \dashv \vdash \neg \paren {p \lor q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|ccccc||cccc|} \hline ...
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_1/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6917
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2
:$\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} }$
{{BeginTableau|\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} } }} {{Assumption|1|\neg p \land \neg q}} {{SequentIntro|2|1|\neg \paren {p \lor q}|1|De Morgan's Laws (Logic): Disjunction of Negations: Formulation 1}} {{Implication|3||\paren {\neg p \land \neg q} \implies \paren {\neg \paren {p \...
:$\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} }$
{{BeginTableau|\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} } }} {{Assumption|1|\neg p \land \neg q}} {{SequentIntro|2|1|\neg \paren {p \lor q}|1|[[De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication|De Morgan's Laws (Logic): Disjunction of Negations: Formulati...
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Proof 1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2/Proof_1
[ "De Morgan's Laws (Logic)" ]
[]
[ "De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Forward Implication", "De Morgan's Laws (Logic)/Disjunction of Negations/Formulation 1/Reverse Implication" ]
proofwiki-6918
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2
:$\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} }$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. :<nowiki>$\begin{array}{|ccccc|c|cccc|} \hline \neg & p & \land & \neg & q & \iff & \neg & (p & \lor & q) \\ \hline \T & \F & \T & \T & \F & \T & \T & \F & \F & \F \\ \T...
:$\vdash \paren {\neg p \land \neg q} \iff \paren {\neg \paren {p \lor q} }$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|...
De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6919
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication
: $p \land q \vdash \neg \left({\neg p \lor \neg q}\right)$
{{BeginTableau|p \land q \vdash \neg \left({\neg p \lor \neg q}\right)}} {{Premise|1|p \land q}} {{Simplification|2|1|p|1|1}} {{Simplification|3|1|q|1|2}} {{Assumption|4|\neg p \lor \neg q}} {{Assumption|5|\neg p}} {{NonContradiction|6|1, 5|2|5}} {{Assumption|7|\neg q}} {{NonContradiction|8|1, 7|3|7}} {{ProofByCases|9|...
: $p \land q \vdash \neg \left({\neg p \lor \neg q}\right)$
{{BeginTableau|p \land q \vdash \neg \left({\neg p \lor \neg q}\right)}} {{Premise|1|p \land q}} {{Simplification|2|1|p|1|1}} {{Simplification|3|1|q|1|2}} {{Assumption|4|\neg p \lor \neg q}} {{Assumption|5|\neg p}} {{NonContradiction|6|1, 5|2|5}} {{Assumption|7|\neg q}} {{NonContradiction|8|1, 7|3|7}} {{ProofByCases|9|...
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_1/Forward_Implication
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6920
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication
: $\neg \paren {\neg p \lor \neg q} \vdash p \land q$
{{BeginTableau|\neg \paren {\neg p \lor \neg q} \vdash p \land q}} {{Premise|1|\neg \paren {\neg p \lor \neg q} }} {{Assumption|2|\neg \paren {p \land q} }} {{DeMorgan|3|2|\neg p \lor \neg q|2|Disjunction of Negations}} {{NonContradiction|4|1, 2|3|1}} {{Reductio|5|1|p \land q|2|4}} {{EndTableau|qed}} {{LEM|Reductio ad ...
: $\neg \paren {\neg p \lor \neg q} \vdash p \land q$
{{BeginTableau|\neg \paren {\neg p \lor \neg q} \vdash p \land q}} {{Premise|1|\neg \paren {\neg p \lor \neg q} }} {{Assumption|2|\neg \paren {p \land q} }} {{DeMorgan|3|2|\neg p \lor \neg q|2|Disjunction of Negations}} {{NonContradiction|4|1, 2|3|1}} {{Reductio|5|1|p \land q|2|4}} {{EndTableau|qed}} {{LEM|Reductio ad ...
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_1/Reverse_Implication
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6921
De Morgan's Laws (Logic)/Conjunction/Formulation 1
:$p \land q \dashv \vdash \neg \paren {\neg p \lor \neg q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||cccccc|} \hline p & \land & q & \neg & (\neg & p & \lor & \neg & q) \\ \hline \F & \F & \F & \F & \T & \F & \T & \T & \F \\ \F & \F & \T & \F & \T &...
:$p \land q \dashv \vdash \neg \paren {\neg p \lor \neg q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccc||cccccc|} \hline p & \land...
De Morgan's Laws (Logic)/Conjunction/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_1/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6922
De Morgan's Laws (Logic)/Conjunction/Formulation 2
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} }$
{{BeginTableau|\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} } }} {{Assumption|1|p \land q}} {{SequentIntro|2|1|\neg \paren {\neg p \lor \neg q}|1|De Morgan's Laws (Logic): Conjunction: Formulation 1}} {{Implication|3||\paren {p \land q} \implies \paren {\neg \paren {\neg p \lor \neg q} }|1|2}...
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} }$
{{BeginTableau|\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} } }} {{Assumption|1|p \land q}} {{SequentIntro|2|1|\neg \paren {\neg p \lor \neg q}|1|[[De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication|De Morgan's Laws (Logic): Conjunction: Formulation 1]]}} {{Implication|3||...
De Morgan's Laws (Logic)/Conjunction/Formulation 2/Proof 1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_2/Proof_1
[ "De Morgan's Laws (Logic)" ]
[]
[ "De Morgan's Laws (Logic)/Conjunction/Formulation 1/Forward Implication", "De Morgan's Laws (Logic)/Conjunction/Formulation 1/Reverse Implication" ]
proofwiki-6923
De Morgan's Laws (Logic)/Conjunction/Formulation 2
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} }$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. $\begin{array}{|ccc|c|cccccc|} \hline (p & \land & q) & \iff & (\neg & (\neg & p & \lor & \neg & q)) \\ \hline \F & \F & \F & \T & \F & \T & \F & \T & \T & \F \\ \F & ...
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {\neg p \lor \neg q} }$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccc|c|cc...
De Morgan's Laws (Logic)/Conjunction/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction/Formulation_2/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6924
De Morgan's Laws (Logic)/Disjunction/Formulation 2
:$\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$
{{BeginTableau|\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} } }} {{Assumption|1|p \lor q}} {{SequentIntro|2|1|\neg \paren {\neg p \land \neg q}|1|De Morgan's Laws (Logic): Disjunction: Formulation 1}} {{Implication|3||\paren {p \lor q} \implies \paren {\neg \paren {\neg p \land \neg q} }|1|2}...
:$\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$
{{BeginTableau|\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} } }} {{Assumption|1|p \lor q}} {{SequentIntro|2|1|\neg \paren {\neg p \land \neg q}|1|[[De Morgan's Laws (Logic)/Disjunction/Formulation 1/Forward Implication|De Morgan's Laws (Logic): Disjunction: Formulation 1]]}} {{Implication|3||...
De Morgan's Laws (Logic)/Disjunction/Formulation 2/Proof 1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_2/Proof_1
[ "De Morgan's Laws (Logic)" ]
[]
[ "De Morgan's Laws (Logic)/Disjunction/Formulation 1/Forward Implication", "De Morgan's Laws (Logic)/Disjunction/Formulation 1/Reverse Implication" ]
proofwiki-6925
De Morgan's Laws (Logic)/Disjunction/Formulation 2
:$\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. :<nowiki>$\begin{array}{|ccc|c|cccccc|} \hline (p & \lor & q) & \iff & (\neg & (\neg & p & \land & \neg & q)) \\ \hline \F & \F & \F & \T & \F & \T & \F & \T & \T & \F ...
:$\vdash \paren {p \lor q} \iff \paren {\neg \paren {\neg p \land \neg q} }$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{...
De Morgan's Laws (Logic)/Disjunction/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_2
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_2/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6926
De Morgan's Laws (Logic)/Disjunction/Formulation 1/Forward Implication
:$p \lor q \vdash \neg \paren {\neg p \land \neg q}$
{{BeginTableau|p \lor q \vdash \neg \paren {\neg p \land \neg q} }} {{Premise|1|p \lor q}} {{Assumption|2|\neg p \land \neg q}} {{Simplification|3|2|\neg p|2|1}} {{Simplification|4|2|\neg q|2|2}} {{Assumption|5|p}} {{NonContradiction|6|2, 5|5|3}} {{Assumption|7|q}} {{NonContradiction|8|2, 7|7|4}} {{ProofByCases|9|1, 2|...
:$p \lor q \vdash \neg \paren {\neg p \land \neg q}$
{{BeginTableau|p \lor q \vdash \neg \paren {\neg p \land \neg q} }} {{Premise|1|p \lor q}} {{Assumption|2|\neg p \land \neg q}} {{Simplification|3|2|\neg p|2|1}} {{Simplification|4|2|\neg q|2|2}} {{Assumption|5|p}} {{NonContradiction|6|2, 5|5|3}} {{Assumption|7|q}} {{NonContradiction|8|2, 7|7|4}} {{ProofByCases|9|1, 2|...
De Morgan's Laws (Logic)/Disjunction/Formulation 1/Forward Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_1/Forward_Implication
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6927
De Morgan's Laws (Logic)/Disjunction/Formulation 1/Reverse Implication
:$\neg \paren {\neg p \land \neg q} \vdash p \lor q$
{{BeginTableau|\neg \paren {\neg p \land \neg q} \vdash p \lor q}} {{Premise|1|\neg \paren {\neg p \land \neg q} }} {{Assumption|2|\neg \paren {p \lor q} }} {{DeMorgan|3|2|\neg p \land \neg q|2|Conjunction of Negations}} {{NonContradiction|4|1, 2|3|1}} {{Reductio|5|1|p \lor q|2|4}} {{EndTableau|qed}} {{LEM|Reductio ad ...
:$\neg \paren {\neg p \land \neg q} \vdash p \lor q$
{{BeginTableau|\neg \paren {\neg p \land \neg q} \vdash p \lor q}} {{Premise|1|\neg \paren {\neg p \land \neg q} }} {{Assumption|2|\neg \paren {p \lor q} }} {{DeMorgan|3|2|\neg p \land \neg q|2|Conjunction of Negations}} {{NonContradiction|4|1, 2|3|1}} {{Reductio|5|1|p \lor q|2|4}} {{EndTableau|qed}} {{LEM|Reductio a...
De Morgan's Laws (Logic)/Disjunction/Formulation 1/Reverse Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_1/Reverse_Implication
[ "De Morgan's Laws (Logic)" ]
[]
[]
proofwiki-6928
De Morgan's Laws (Logic)/Disjunction/Formulation 1
:$p \lor q \dashv \vdash \neg \paren {\neg p \land \neg q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||cccccc|} \hline p & \lor & q & \neg & (\neg & p & \land & \neg & q) \\ \hline \F & \F & \F & \F & \T & \F & \T & \T & \F \\ \F & \T & \T & \T & \T & ...
:$p \lor q \dashv \vdash \neg \paren {\neg p \land \neg q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccc||cccccc|} \hline p & \lor ...
De Morgan's Laws (Logic)/Disjunction/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_1
https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Disjunction/Formulation_1/Proof_by_Truth_Table
[ "De Morgan's Laws (Logic)" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6929
Requirement for Connected Domain to be Simply Connected Domain
Then $D$ is '''simply connected''' {{iff}} all paths in $D$ with the same initial points and end points are path-homotopic. That is, given two paths in $D$: :$\gamma: \closedint a b \to D$ :$\sigma: \closedint c d \to D$ with $\map \gamma a = \map \sigma c$ and $\map \gamma b = \map \sigma d$, there exists a continuous...
From Open Domain is Connected iff it is Path-Connected, it follows that $D$ is path-connected. The results now follows by the definition of simply connected sets. {{qed}}
Then $D$ is '''simply connected''' {{iff}} all [[Definition:Path (Topology)|paths]] in $D$ with the same [[Definition:Initial Point of Path|initial points]] and [[Definition:Endpoint of Path|end points]] are [[Definition:Path-Homotopic|path-homotopic]]. That is, given two [[Definition:Path (Topology)|paths]] in $D$: ...
From [[Open Domain is Connected iff it is Path-Connected]], it follows that $D$ is [[Definition:Path-Connected|path-connected]]. The results now follows by the definition of [[Definition:Simply Connected/Definition 3|simply connected sets]]. {{qed}}
Requirement for Connected Domain to be Simply Connected Domain
https://proofwiki.org/wiki/Requirement_for_Connected_Domain_to_be_Simply_Connected_Domain
https://proofwiki.org/wiki/Requirement_for_Connected_Domain_to_be_Simply_Connected_Domain
[ "Complex Analysis", "Simply Connected Spaces" ]
[ "Definition:Path (Topology)", "Definition:Path (Topology)/Initial Point", "Definition:Path (Topology)/Endpoint", "Definition:Homotopy/Path", "Definition:Path (Topology)", "Definition:Continuous Complex Function" ]
[ "Open Domain is Connected iff it is Path-Connected", "Definition:Path-Connected", "Definition:Simply Connected/Definition 3" ]
proofwiki-6930
Rule of Distribution/Disjunction Distributes over Conjunction
=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline p & \lor & (q & \land & r) & (p & \lor & q) & \land & (p & \lor & r) \\ \hline \F & \F & \F & \F & \F & \F & \F...
=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive|Disjunction is Left Distributive over Conjunction]] === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive|Disj...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_1/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive", "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6931
Rule of Distribution/Disjunction Distributes over Conjunction
=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|instance 1 of a Gentzen proof system}} {{TableauLine | n = 1 | f = \neg p, p, q | rlnk = Definition:Gentzen Proof System/Instance 1#Axioms | rtxt = Axiom...
=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive|Disjunction is Left Distributive over Conjunction]] === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive|Disj...
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|[[Definition:Gentzen Proof System/Instance 1|instance 1 of a Gentzen proof system]]}} {{TableauLine | n = 1 | f = \neg p, p, q | rlnk = Definition:Gentzen Proof System...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Forward_Implication/Proof_1
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive", "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive" ]
[ "Definition:Gentzen Proof System/Instance 1" ]
proofwiki-6932
Rule of Distribution/Disjunction Distributes over Conjunction
=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|Instance 2 of the Hilbert-style systems}} {{TableauLine | n = 1 | f = \paren {q \land r} \implies q | rlnk = Rule of Simplification/Sequent Form/Formulation 2/Form 1 | rtxt = Rule of Simplification...
=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive|Disjunction is Left Distributive over Conjunction]] === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive|Disj...
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}} {{TableauLine | n = 1 | f = \paren {q \land r} \implies q | rlnk = Rule of Simplification/Sequent Form/Formul...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication/Proof 2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Forward_Implication/Proof_2
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive", "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive" ]
[ "Definition:Hilbert Proof System/Instance 2" ]
proofwiki-6933
Rule of Distribution/Disjunction Distributes over Conjunction
=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} } }} {{Assumption|1|p \lor \paren {q \land r} }} {{SequentIntro|2|1|\paren {p \lor q} \land \paren {p \lor r}|1|Disjunction is Left Distributive over Conjunction: Formulation 1}} {{Implication|3||\paren {p \...
=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive|Disjunction is Left Distributive over Conjunction]] === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive|Disj...
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} } }} {{Assumption|1|p \lor \paren {q \land r} }} {{SequentIntro|2|1|\paren {p \lor q} \land \paren {p \lor r}|1|[[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Proof_1
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive", "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Forward Implication", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Reverse Implication" ]
proofwiki-6934
Rule of Distribution/Disjunction Distributes over Conjunction
=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \lor & (q & \land & r) & \iff & (p & \lor & q) & \land & (p & \lor & r) \\ \hline \F & \F & \F & \F & \F ...
=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive|Disjunction is Left Distributive over Conjunction]] === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive|Disj...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begi...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive", "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6935
Rule of Distribution/Disjunction Distributes over Conjunction
=== Disjunction is Left Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === Disjunction is Right Distributive over Conjunction === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline (q & \land & r) & \lor & p & (q & \lor & p) & \land & (r & \lor & p) \\ \hline F & F & F & F & F & F & F & F & ...
=== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive|Disjunction is Left Distributive over Conjunction]] === {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive}} === [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive|Disj...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1/Proof
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_1/Proof
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive", "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6936
Rule of Distribution/Conjunction Distributes over Disjunction
=== Conjunction is Left Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === Conjunction is Right Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline p & \land & (q & \lor & r) & (p & \land & q) & \lor & (p & \land & r) \\ \hline \F & \F & \F & \F & \F & \F & \...
=== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive|Conjunction is Left Distributive over Disjunction]] === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive|Conju...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_1/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6937
Rule of Distribution/Conjunction Distributes over Disjunction
=== Conjunction is Left Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === Conjunction is Right Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{Assumption|1|p \land \paren {q \lor r} }} {{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|Conjunction is Left Distributive over Disjunction: Formulation 1}} {{Implication|3|...
=== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive|Conjunction is Left Distributive over Disjunction]] === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive|Conju...
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{Assumption|1|p \land \paren {q \lor r} }} {{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/F...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_1
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Forward Implication", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Reverse Implication" ]
proofwiki-6938
Rule of Distribution/Conjunction Distributes over Disjunction
=== Conjunction is Left Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === Conjunction is Right Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{TheoremIntro|1|\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }|Conjunction Distributes over Disjunction: Forward Implication}} {{TheoremIntro|2|\pare...
=== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive|Conjunction is Left Distributive over Disjunction]] === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive|Conju...
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{TheoremIntro|1|\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Form...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_2
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Reverse Implication" ]
proofwiki-6939
Rule of Distribution/Conjunction Distributes over Disjunction
=== Conjunction is Left Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === Conjunction is Right Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \land & (q & \lor & r) & \iff & (p & \land & q) & \lor & (p & \land & r) \\ \hline \F & \F & \F & \F & \F...
=== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive|Conjunction is Left Distributive over Disjunction]] === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive|Conju...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begi...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6940
Rule of Distribution/Conjunction Distributes over Disjunction
=== Conjunction is Left Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === Conjunction is Right Distributive over Disjunction === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :<nowiki>$\begin{array}{|ccccc||ccccccc|} \hline (q & \lor & r) & \land & p & (q & \land & p) & \lor & (r & \land & p) \\ \hline \F & \F & \F & \F & \F...
=== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive|Conjunction is Left Distributive over Disjunction]] === {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive}} === [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive|Conju...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|cc...
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Right_Distributive/Formulation_1/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6941
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{Assumption|1|p \land \paren {q \lor r} }} {{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|Conjunction is Left Distributive over Disjunction: Formulation 1}} {{Implication|3|...
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{Assumption|1|p \land \paren {q \lor r} }} {{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/F...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_1
[ "Rule of Distribution" ]
[]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Forward Implication", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Reverse Implication" ]
proofwiki-6942
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{TheoremIntro|1|\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }|Conjunction Distributes over Disjunction: Forward Implication}} {{TheoremIntro|2|\pare...
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{TheoremIntro|1|\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Form...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_2
[ "Rule of Distribution" ]
[]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Reverse Implication" ]
proofwiki-6943
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \land & (q & \lor & r) & \iff & (p & \land & q) & \lor & (p & \land & r) \\ \hline \F & \F & \F & \F & \F...
:$\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }$
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begi...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6944
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline p & \land & (q & \lor & r) & (p & \land & q) & \lor & (p & \land & r) \\ \hline \F & \F & \F & \F & \F & \F & \...
==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_1/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6945
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}}
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{Assumption|1|p \land \paren {q \lor r} }} {{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|Conjunction is Left Distributive over Disjunction: Formulation 1}} {{Implication|3|...
==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation...
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{Assumption|1|p \land \paren {q \lor r} }} {{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/F...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_1
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Forward Implication", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1/Reverse Implication" ]
proofwiki-6946
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}}
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{TheoremIntro|1|\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }|Conjunction Distributes over Disjunction: Forward Implication}} {{TheoremIntro|2|\pare...
==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation...
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{TheoremIntro|1|\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }|[[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Form...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof 2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_2
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Forward Implication", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Reverse Implication" ]
proofwiki-6947
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \land & (q & \lor & r) & \iff & (p & \land & q) & \lor & (p & \land & r) \\ \hline \F & \F & \F & \F & \F...
==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begi...
Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Left_Distributive/Formulation_2/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 1", "Rule of Distribution/Conjunction Distributes over Disjunction/Left Distributive/Formulation 2" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6948
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :<nowiki>$\begin{array}{|ccccc||ccccccc|} \hline (q & \lor & r) & \land & p & (q & \land & p) & \lor & (r & \land & p) \\ \hline \F & \F & \F & \F & \F...
==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1}} ==== [[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulat...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|cc...
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Right_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Right_Distributive/Formulation_1/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6949
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1
:$\paren {q \lor r} \land p \dashv \vdash \paren {q \land p} \lor \paren {r \land p}$
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :<nowiki>$\begin{array}{|ccccc||ccccccc|} \hline (q & \lor & r) & \land & p & (q & \land & p) & \lor & (r & \land & p) \\ \hline \F & \F & \F & \F & \F...
:$\paren {q \lor r} \land p \dashv \vdash \paren {q \land p} \lor \paren {r \land p}$
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|cc...
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Right_Distributive/Formulation_1
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Right_Distributive/Formulation_1/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6950
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2
:$\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }$
{{BeginTableau|\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} } }} {{Assumption|1|\paren {q \lor r} \land p}} {{SequentIntro|2|1|\paren {q \land p} \lor \paren {r \land p}|1|Conjunction is Right Distributive over Disjunction: Formulation 1}} {{Implication|3||\paren {\p...
:$\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} }$
{{BeginTableau|\vdash \paren {\paren {q \lor r} \land p} \iff \paren {\paren {q \land p} \lor \paren {r \land p} } }} {{Assumption|1|\paren {q \lor r} \land p}} {{SequentIntro|2|1|\paren {q \land p} \lor \paren {r \land p}|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation ...
Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Right_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Conjunction_Distributes_over_Disjunction/Right_Distributive/Formulation_2
[ "Rule of Distribution" ]
[]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1", "Category:Rule of Distribution" ]
proofwiki-6951
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline p & \lor & (q & \land & r) & (p & \lor & q) & \land & (p & \lor & r) \\ \hline \F & \F & \F & \F & \F & \F & \F...
==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation ...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_1/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6952
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|instance 1 of a Gentzen proof system}} {{TableauLine | n = 1 | f = \neg p, p, q | rlnk = Definition:Gentzen Proof System/Instance 1#Axioms | rtxt = Axiom...
==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation ...
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|[[Definition:Gentzen Proof System/Instance 1|instance 1 of a Gentzen proof system]]}} {{TableauLine | n = 1 | f = \neg p, p, q | rlnk = Definition:Gentzen Proof System...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Forward_Implication/Proof_1
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2" ]
[ "Definition:Gentzen Proof System/Instance 1" ]
proofwiki-6953
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|Instance 2 of the Hilbert-style systems}} {{TableauLine | n = 1 | f = \paren {q \land r} \implies q | rlnk = Rule of Simplification/Sequent Form/Formulation 2/Form 1 | rtxt = Rule of Simplification...
==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation ...
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}} {{TableauLine | n = 1 | f = \paren {q \land r} \implies q | rlnk = Rule of Simplification/Sequent Form/Formul...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication/Proof 2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Forward_Implication/Proof_2
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2" ]
[ "Definition:Hilbert Proof System/Instance 2" ]
proofwiki-6954
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} } }} {{Assumption|1|p \lor \paren {q \land r} }} {{SequentIntro|2|1|\paren {p \lor q} \land \paren {p \lor r}|1|Disjunction is Left Distributive over Conjunction: Formulation 1}} {{Implication|3||\paren {p \...
==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation ...
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} } }} {{Assumption|1|p \lor \paren {q \land r} }} {{SequentIntro|2|1|\paren {p \lor q} \land \paren {p \lor r}|1|[[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Proof_1
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Forward Implication", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Reverse Implication" ]
proofwiki-6955
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \lor & (q & \land & r) & \iff & (p & \lor & q) & \land & (p & \lor & r) \\ \hline \F & \F & \F & \F & \F ...
==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1}} ==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation ...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begi...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6956
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive
==== Formulation 1 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1}} ==== Formulation 2 ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2}}
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline (q & \land & r) & \lor & p & (q & \lor & p) & \land & (r & \lor & p) \\ \hline F & F & F & F & F & F & F & F & ...
==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1|Formulation 1]] ==== {{:Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1}} ==== [[Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulati...
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1/Proof
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_1/Proof
[ "Rule of Distribution" ]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1", "Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2" ]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6957
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Forward Implication
:$p \lor \paren {q \land r} \vdash \paren {p \lor q} \land \paren {p \lor r}$
{{BeginTableau|p \lor \paren {q \land r} \vdash \paren {p \lor q} \land \paren {p \lor r} }} {{Premise | 1|p \lor \paren {q \land r} }} {{Assumption | 2|p}} {{Addition | 3|2|p \lor q|2|1}} {{Addition | 4|2|p \lor r|2|1}} {{Conjunction | 5|2|\paren {p \lor q} \land \paren {p \lor r}|3|4}} {{Ass...
:$p \lor \paren {q \land r} \vdash \paren {p \lor q} \land \paren {p \lor r}$
{{BeginTableau|p \lor \paren {q \land r} \vdash \paren {p \lor q} \land \paren {p \lor r} }} {{Premise | 1|p \lor \paren {q \land r} }} {{Assumption | 2|p}} {{Addition | 3|2|p \lor q|2|1}} {{Addition | 4|2|p \lor r|2|1}} {{Conjunction | 5|2|\paren {p \lor q} \land \paren {p \lor r}|3|4}} {{Ass...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Forward Implication
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_1/Forward_Implication
[ "Rule of Distribution" ]
[]
[ "Category:Rule of Distribution" ]
proofwiki-6958
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|instance 1 of a Gentzen proof system}} {{TableauLine | n = 1 | f = \neg p, p, q | rlnk = Definition:Gentzen Proof System/Instance 1#Axioms | rtxt = Axiom...
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|[[Definition:Gentzen Proof System/Instance 1|instance 1 of a Gentzen proof system]]}} {{TableauLine | n = 1 | f = \neg p, p, q | rlnk = Definition:Gentzen Proof System...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Forward_Implication/Proof_1
[ "Rule of Distribution" ]
[]
[ "Definition:Gentzen Proof System/Instance 1" ]
proofwiki-6959
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|Instance 2 of the Hilbert-style systems}} {{TableauLine | n = 1 | f = \paren {q \land r} \implies q | rlnk = Rule of Simplification/Sequent Form/Formulation 2/Form 1 | rtxt = Rule of Simplification...
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \implies \paren {\paren {p \lor q} \land \paren {p \lor r} }|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}} {{TableauLine | n = 1 | f = \paren {q \land r} \implies q | rlnk = Rule of Simplification/Sequent Form/Formul...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Forward Implication/Proof 2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Forward_Implication/Proof_2
[ "Rule of Distribution" ]
[]
[ "Definition:Hilbert Proof System/Instance 2" ]
proofwiki-6960
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} } }} {{Assumption|1|p \lor \paren {q \land r} }} {{SequentIntro|2|1|\paren {p \lor q} \land \paren {p \lor r}|1|Disjunction is Left Distributive over Conjunction: Formulation 1}} {{Implication|3||\paren {p \...
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
{{BeginTableau|\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} } }} {{Assumption|1|p \lor \paren {q \land r} }} {{SequentIntro|2|1|\paren {p \lor q} \land \paren {p \lor r}|1|[[Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Proof 1
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Proof_1
[ "Rule of Distribution" ]
[]
[ "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Forward Implication", "Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 1/Reverse Implication" ]
proofwiki-6961
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. $\begin{array}{|ccccc|c|ccccccc|} \hline p & \lor & (q & \land & r) & \iff & (p & \lor & q) & \land & (p & \lor & r) \\ \hline \F & \F & \F & \F & \F ...
:$\vdash \paren {p \lor \paren {q \land r} } \iff \paren {\paren {p \lor q} \land \paren {p \lor r} }$
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begi...
Rule of Distribution/Disjunction Distributes over Conjunction/Left Distributive/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Left_Distributive/Formulation_2/Proof_by_Truth_Table
[ "Rule of Distribution" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6962
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1
:$\paren {q \land r} \lor p \dashv \vdash \paren {q \lor p} \land \paren {r \lor p}$
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline (q & \land & r) & \lor & p & (q & \lor & p) & \land & (r & \lor & p) \\ \hline F & F & F & F & F & F & F & F & ...
:$\paren {q \land r} \lor p \dashv \vdash \paren {q \lor p} \land \paren {r \lor p}$
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1/Proof
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_1
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_1/Proof
[ "Rule of Distribution" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6963
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2
:$\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }$
{{BeginTableau|\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} } }} {{Assumption|1|\paren {q \land r} \lor p}} {{SequentIntro|2|1|\paren {q \lor p} \land \paren {r \lor p}|1|Conjunction is Right Distributive over Disjunction: Formulation 1}} {{Implication|3||\paren {\par...
:$\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} }$
{{BeginTableau|\vdash \paren {\paren {q \land r} \lor p} \iff \paren {\paren {q \lor p} \land \paren {r \lor p} } }} {{Assumption|1|\paren {q \land r} \lor p}} {{SequentIntro|2|1|\paren {q \lor p} \land \paren {r \lor p}|1|[[Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1|...
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_2
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_2
[ "Rule of Distribution" ]
[]
[ "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1", "Rule of Distribution/Conjunction Distributes over Disjunction/Right Distributive/Formulation 1", "Category:Rule of Distribution" ]
proofwiki-6964
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1/Proof
:$\left({q \land r}\right) \lor p \dashv \vdash \left({q \lor p}\right) \land \left({r \lor p}\right)$
We apply the Method of Truth Tables to the proposition. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccccc||ccccccc|} \hline (q & \land & r) & \lor & p & (q & \lor & p) & \land & (r & \lor & p) \\ \hline F & F & F & F & F & F & F & F & ...
:$\left({q \land r}\right) \lor p \dashv \vdash \left({q \lor p}\right) \land \left({r \lor p}\right)$
We apply the [[Method of Truth Tables]] to the proposition. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccccc||cccc...
Rule of Distribution/Disjunction Distributes over Conjunction/Right Distributive/Formulation 1/Proof
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_1/Proof
https://proofwiki.org/wiki/Rule_of_Distribution/Disjunction_Distributes_over_Conjunction/Right_Distributive/Formulation_1/Proof
[ "Rule of Distribution", "Truth Table Proofs" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6965
Union of Subsets is Subset/Set of Sets
Let $T$ be a set. Let $\mathbb S$ be a set of sets. Suppose that for each $S \in \mathbb S$, $S \subseteq T$. Then: :$\ds \bigcup \mathbb S \subseteq T$
Let $x \in \ds \bigcup \mathbb S$. By the definition of union, there exists an $S \in \mathbb S$ such that $x \in S$. By premise, $S \subseteq T$. By the definition of subset, $x \in T$. Since this result holds for each $x \in \ds \bigcup \mathbb S$: :$\ds \bigcup \mathbb S \subseteq T$ {{qed}} Category:Union of Subset...
Let $T$ be a [[Definition:Set|set]]. Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]]. Suppose that for each $S \in \mathbb S$, $S \subseteq T$. Then: :$\ds \bigcup \mathbb S \subseteq T$
Let $x \in \ds \bigcup \mathbb S$. By the definition of [[Definition:Set Union|union]], there exists an $S \in \mathbb S$ such that $x \in S$. By premise, $S \subseteq T$. By the definition of [[Definition:Subset|subset]], $x \in T$. Since this result holds for each $x \in \ds \bigcup \mathbb S$: :$\ds \bigcup \ma...
Union of Subsets is Subset/Set of Sets
https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Set_of_Sets
https://proofwiki.org/wiki/Union_of_Subsets_is_Subset/Set_of_Sets
[ "Union of Subsets is Subset" ]
[ "Definition:Set", "Definition:Set of Sets" ]
[ "Definition:Set Union", "Definition:Subset", "Category:Union of Subsets is Subset" ]
proofwiki-6966
Union is Smallest Superset/Set of Sets
Let $T$ be a set. Let $\mathbb S$ be a set of sets. Then: :$\paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
By Union of Subsets is Subset: Set of Sets: :$\paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$ {{qed|lemma}} For the converse implication, suppose that $\bigcup \mathbb S \subseteq T$. Consider any $X \in \mathbb S$ and take any $x \in X$. From Set is Subset of Union: Set of Sets ...
Let $T$ be a [[Definition:Set|set]]. Let $\mathbb S$ be a [[Definition:Set of Sets|set of sets]]. Then: :$\paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$
By [[Union of Subsets is Subset/Set of Sets|Union of Subsets is Subset: Set of Sets]]: :$\paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$ {{qed|lemma}} For the converse implication, suppose that $\bigcup \mathbb S \subseteq T$. Consider any $X \in \mathbb S$ and take any $x \...
Union is Smallest Superset/Set of Sets
https://proofwiki.org/wiki/Union_is_Smallest_Superset/Set_of_Sets
https://proofwiki.org/wiki/Union_is_Smallest_Superset/Set_of_Sets
[ "Set Union", "Subsets" ]
[ "Definition:Set", "Definition:Set of Sets" ]
[ "Union of Subsets is Subset/Set of Sets", "Set is Subset of Union/Set of Sets" ]
proofwiki-6967
Intersection of Relations is Relation
Let $S$ and $T$ be sets. Let $\FF$ be a family of relations from $S$ to $T$. Let $\ds \RR = \bigcap \FF$, the intersection of all the elements of $\FF$. Then $\RR$ is a relation from $S$ to $T$. {{Expand|Binary case}}
By the definition of a relation from $S$ to $T$, each element of $\FF$ is a subset of $S \times T$. By Intersection of Subsets is Subset: Set of Sets: :$\RR \subseteq S \times T$ Therefore, by the definition of a relation from $S$ to $T$, $\RR$ is a relation from $S$ to $T$. {{qed}} Category:Relation Theory fgpu3jpko1f...
Let $S$ and $T$ be [[Definition:Set|sets]]. Let $\FF$ be a family of [[Definition:Relation|relations]] from $S$ to $T$. Let $\ds \RR = \bigcap \FF$, the [[Definition:Intersection of Set of Sets|intersection]] of all the [[Definition:Element|elements]] of $\FF$. Then $\RR$ is a [[Definition:Relation|relation]] from...
By the definition of a [[Definition:Relation|relation]] from $S$ to $T$, each [[Definition:Element|element]] of $\FF$ is a [[Definition:Subset|subset]] of $S \times T$. By [[Intersection of Subsets is Subset/Set of Sets|Intersection of Subsets is Subset: Set of Sets]]: :$\RR \subseteq S \times T$ Therefore, by the d...
Intersection of Relations is Relation
https://proofwiki.org/wiki/Intersection_of_Relations_is_Relation
https://proofwiki.org/wiki/Intersection_of_Relations_is_Relation
[ "Relation Theory" ]
[ "Definition:Set", "Definition:Relation", "Definition:Set Intersection/Set of Sets", "Definition:Element", "Definition:Relation" ]
[ "Definition:Relation", "Definition:Element", "Definition:Subset", "Intersection of Subsets is Subset/Set of Sets", "Definition:Relation", "Definition:Relation", "Category:Relation Theory" ]
proofwiki-6968
Intersection of Subsets is Subset/Set of Sets
Let $T$ be a set. Let $\mathbb S$ be a non-empty set of sets. Suppose that for each $S \in \mathbb S$: :$S \subseteq T$ Then: :$\bigcap \mathbb S \subseteq T$
Let $x \in \bigcap \mathbb S$. Then by the definition of intersection: :$\forall S \in \mathbb S: x \in S$. Since $\mathbb S$ is non-empty by the premise, it has some element $S$. Then $x \in S$. Since $S \in \mathbb S$, the premise shows that $S \subseteq T$. By the definition of subset, $x \in T$. Since this holds fo...
Let $T$ be a [[Definition:Set|set]]. Let $\mathbb S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set of sets]]. Suppose that for each $S \in \mathbb S$: :$S \subseteq T$ Then: :$\bigcap \mathbb S \subseteq T$
Let $x \in \bigcap \mathbb S$. Then by the definition of [[Definition:Set Intersection|intersection]]: :$\forall S \in \mathbb S: x \in S$. Since $\mathbb S$ is [[Definition:Non-Empty Set|non-empty]] by the premise, it has some [[Definition:Element|element]] $S$. Then $x \in S$. Since $S \in \mathbb S$, the premise...
Intersection of Subsets is Subset/Set of Sets
https://proofwiki.org/wiki/Intersection_of_Subsets_is_Subset/Set_of_Sets
https://proofwiki.org/wiki/Intersection_of_Subsets_is_Subset/Set_of_Sets
[ "Set Intersection", "Subsets" ]
[ "Definition:Set", "Definition:Non-Empty Set", "Definition:Set of Sets" ]
[ "Definition:Set Intersection", "Definition:Non-Empty Set", "Definition:Element", "Definition:Subset", "Category:Set Intersection", "Category:Subsets" ]
proofwiki-6969
Equivalence of Definitions of Initial Topology
Let $X$ be a set. Let $I$ be an indexing set. Let $\family {\struct{Y_i, \tau_i}}_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$. Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$. {{TFAE|def = Initial Topology}}
=== Definition 1 Implies Definition 2 === {{:Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2}}{{qed|lemma}}
Let $X$ be a [[Definition:Set|set]]. Let $I$ be an [[Definition:Indexing Set|indexing set]]. Let $\family {\struct{Y_i, \tau_i}}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] [[Definition:Indexing Set|indexed]] by $I$. Let $\family {f_i:...
=== [[Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2|Definition 1 Implies Definition 2]] === {{:Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2}}{{qed|lemma}}
Equivalence of Definitions of Initial Topology
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Initial_Topology
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Initial_Topology
[ "Initial Topology" ]
[ "Definition:Set", "Definition:Indexing Set", "Definition:Indexing Set/Family", "Definition:Topological Space", "Definition:Indexing Set", "Definition:Indexing Set/Family", "Definition:Mapping", "Definition:Indexing Set" ]
[ "Equivalence of Definitions of Initial Topology/Definition 1 Implies Definition 2" ]
proofwiki-6970
Radius of Convergence from Limit of Sequence/Complex Case
Let $\xi \in \C$ be a complex number. Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$. Let the sequence $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converge. Then $R$ is given by: :$\ds \dfrac 1 R = \lim_{n \mathop \to \infty} \cmo...
Let the sequence $\sequence {\cmod {\dfrac {a_{n+1} } {a_n} } }_{n \mathop \in \N}$ converge. Let $\epsilon \in \R_{>0}$, and let $z \in \C$. Let $\cmod {z - \xi} = R - \epsilon$. By definition of radius of convergence, it follows that $\map S z$ is absolutely convergent. Then from the Ratio Test: :$\lim_{n \mathop \to...
Let $\xi \in \C$ be a [[Definition:Complex Number|complex number]]. Let $\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a [[Definition:Complex Power Series|(complex) power series]] about $\xi$. Let the [[Definition:Infinite Sequence|sequence]] $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{...
Let the [[Definition:Infinite Sequence|sequence]] $\sequence {\cmod {\dfrac {a_{n+1} } {a_n} } }_{n \mathop \in \N}$ [[Definition:Convergent Real Sequence|converge]]. Let $\epsilon \in \R_{>0}$, and let $z \in \C$. Let $\cmod {z - \xi} = R - \epsilon$. By definition of [[Definition:Radius of Convergence of Complex P...
Radius of Convergence from Limit of Sequence/Complex Case
https://proofwiki.org/wiki/Radius_of_Convergence_from_Limit_of_Sequence/Complex_Case
https://proofwiki.org/wiki/Radius_of_Convergence_from_Limit_of_Sequence/Complex_Case
[ "Radius of Convergence from Limit of Sequence", "Complex Power Series" ]
[ "Definition:Complex Number", "Definition:Power Series/Complex Domain", "Definition:Sequence/Infinite Sequence", "Definition:Convergent Sequence/Analysis", "Definition:Absolutely Convergent Series" ]
[ "Definition:Sequence/Infinite Sequence", "Definition:Convergent Sequence/Real Numbers", "Definition:Radius of Convergence/Complex Domain", "Definition:Absolutely Convergent Series", "Ratio Test", "Combination Theorem for Sequences/Complex/Multiple Rule", "Definition:Divergent Series", "Ratio Test" ]
proofwiki-6971
Radius of Convergence from Limit of Sequence/Real Case
Let $\xi \in \R$ be a real number. Let $\ds S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$. Then the radius of convergence $R$ of $S \paren x$ is given by: :$\ds \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$ if this limit exists and is nonzero...
From the ratio test, $S \paren x$ is convergent if: :$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } < 1$ Thus: {{begin-eqn}} {{eqn | l = \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } | o = < | r = 1 | c = ...
Let $\xi \in \R$ be a [[Definition:Real Number|real number]]. Let $\ds S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a [[Definition:Power Series|power series]] about $\xi$. Then the [[Definition:Radius of Convergence|radius of convergence]] $R$ of $S \paren x$ is given by: :$\ds \frac 1 R = \li...
From the [[Ratio Test|ratio test]], $S \paren x$ is [[Definition:Convergent Series|convergent]] if: :$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \xi}^n} } < 1$ Thus: {{begin-eqn}} {{eqn | l = \size {\frac {a_{n + 1} \paren {x - \xi}^{n + 1} } {a_n \paren {x - \...
Radius of Convergence from Limit of Sequence/Real Case
https://proofwiki.org/wiki/Radius_of_Convergence_from_Limit_of_Sequence/Real_Case
https://proofwiki.org/wiki/Radius_of_Convergence_from_Limit_of_Sequence/Real_Case
[ "Radius of Convergence from Limit of Sequence", "Real Power Series" ]
[ "Definition:Real Number", "Definition:Power Series", "Definition:Radius of Convergence", "Definition:Radius of Convergence", "Definition:Interval of Convergence" ]
[ "Ratio Test", "Definition:Convergent Series", "Definition:Radius of Convergence" ]
proofwiki-6972
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 1
:$p \land q \dashv \vdash \neg \paren {p \implies \neg q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|ccc||ccccc|} \hline p & \land & q & \neg & (p & \implies & \neg & q) \\ \hline \F & \F & \F & \F & \F & \T & \T & \F \\ \F & \F & \T & \F & \F & \T & \F &...
:$p \land q \dashv \vdash \neg \paren {p \implies \neg q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. $\begin{array}{|ccc||ccccc|} \hline p & \land ...
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_1
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_1/Proof_by_Truth_Table
[ "Conjunction is Equivalent to Negation of Conditional of Negative" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6973
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 2
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$
{{BeginTableau|\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} } }} {{Assumption|1|p \land q}} {{SequentIntro|2|1|\neg \paren {p \implies \neg q}|1|Conjunction is Equivalent to Negation of Conditional of Negative: Formulation 1: Forward Implication}} {{Implication|3||\paren {p \land q} \implies \...
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$
{{BeginTableau|\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} } }} {{Assumption|1|p \land q}} {{SequentIntro|2|1|\neg \paren {p \implies \neg q}|1|[[Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 1/Forward Implication|Conjunction is Equivalent to Negation of Conditi...
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 2/Proof 1
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_2
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_2/Proof_1
[ "Conjunction is Equivalent to Negation of Conditional of Negative" ]
[]
[ "Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 1/Forward Implication", "Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 1/Reverse Implication" ]
proofwiki-6974
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 2
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. :<nowiki>$\begin{array}{|ccc|c|ccccc|} \hline (p & \land & q) & \iff & (\neg & (p & \implies & \neg & q)) \\ \hline \F & \F & \F & \T & \F & \F & \T & \T & \F \\ \F & \...
:$\vdash \paren {p \land q} \iff \paren {\neg \paren {p \implies \neg q} }$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{...
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_2
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_2/Proof_by_Truth_Table
[ "Conjunction is Equivalent to Negation of Conditional of Negative" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6975
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 1/Forward Implication
:$p \land q \vdash \neg \paren {p \implies \neg q}$
{{BeginTableau|p \land q \vdash \neg \paren {p \implies \neg q} }} {{Premise|1|p \land q}} {{Assumption|2|p \implies \neg q|Assume the opposite of what is to be proved ...}} {{Simplification|3|1|p|1|1}} {{Simplification|4|1|q|1|2}} {{ModusPonens|5|1, 2|\neg q|2|3}} {{NonContradiction|6|1, 2|4|5| ... and demonstrate a c...
:$p \land q \vdash \neg \paren {p \implies \neg q}$
{{BeginTableau|p \land q \vdash \neg \paren {p \implies \neg q} }} {{Premise|1|p \land q}} {{Assumption|2|p \implies \neg q|Assume the opposite of what is to be proved ...}} {{Simplification|3|1|p|1|1}} {{Simplification|4|1|q|1|2}} {{ModusPonens|5|1, 2|\neg q|2|3}} {{NonContradiction|6|1, 2|4|5| ... and demonstrate a c...
Conjunction is Equivalent to Negation of Conditional of Negative/Formulation 1/Forward Implication
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/Conjunction_is_Equivalent_to_Negation_of_Conditional_of_Negative/Formulation_1/Forward_Implication
[ "Conjunction is Equivalent to Negation of Conditional of Negative" ]
[]
[ "Category:Conjunction is Equivalent to Negation of Conditional of Negative" ]
proofwiki-6976
Rule of Material Implication/Formulation 1/Forward Implication
:$p \implies q \vdash \neg p \lor q$
{{BeginTableau|p \implies q \vdash \neg p \lor q}} {{Premise|1|p \implies q}} {{ExcludedMiddle|2|p \lor \neg p}} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor q|3|1}} {{Assumption|5|p}} {{ModusPonens|6|1, 5|q|1|5}} {{Addition|7|1, 5|\neg p \lor q|6|2}} {{ProofByCases|8|1|\neg p \lor q|2|3|4|5|7}} {{EndTableau}} {{...
:$p \implies q \vdash \neg p \lor q$
{{BeginTableau|p \implies q \vdash \neg p \lor q}} {{Premise|1|p \implies q}} {{ExcludedMiddle|2|p \lor \neg p}} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor q|3|1}} {{Assumption|5|p}} {{ModusPonens|6|1, 5|q|1|5}} {{Addition|7|1, 5|\neg p \lor q|6|2}} {{ProofByCases|8|1|\neg p \lor q|2|3|4|5|7}} {{EndTableau}} {{...
Rule of Material Implication/Formulation 1/Forward Implication/Proof
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Forward_Implication/Proof
[ "Rule of Material Implication" ]
[]
[]
proofwiki-6977
Rule of Material Implication/Formulation 1
:$p \implies q \dashv \vdash \neg p \lor q$
{{BeginTableau|p \implies q \vdash \neg p \lor q}} {{Premise|1|p \implies q}} {{ExcludedMiddle|2|p \lor \neg p}} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor q|3|1}} {{Assumption|5|p}} {{ModusPonens|6|1, 5|q|1|5}} {{Addition|7|1, 5|\neg p \lor q|6|2}} {{ProofByCases|8|1|\neg p \lor q|2|3|4|5|7}} {{EndTableau}} {{...
:$p \implies q \dashv \vdash \neg p \lor q$
{{BeginTableau|p \implies q \vdash \neg p \lor q}} {{Premise|1|p \implies q}} {{ExcludedMiddle|2|p \lor \neg p}} {{Assumption|3|\neg p}} {{Addition|4|3|\neg p \lor q|3|1}} {{Assumption|5|p}} {{ModusPonens|6|1, 5|q|1|5}} {{Addition|7|1, 5|\neg p \lor q|6|2}} {{ProofByCases|8|1|\neg p \lor q|2|3|4|5|7}} {{EndTableau}} {{...
Rule of Material Implication/Formulation 1/Forward Implication/Proof
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Forward_Implication/Proof
[ "Rule of Material Implication" ]
[]
[]
proofwiki-6978
Rule of Material Implication/Formulation 1
:$p \implies q \dashv \vdash \neg p \lor q$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :<nowiki>$\begin {array} {|ccc||cccc|} \hline p & \implies & q & \neg & p & \lor & q \\ \hline \F & \T & \F & \T & \F & \T & \F \\ \F & \T & \T & \T & \F & \T & \T \\ \T ...
:$p \implies q \dashv \vdash \neg p \lor q$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin {array} {|ccc||cccc|} \hline ...
Rule of Material Implication/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Proof_by_Truth_Table
[ "Rule of Material Implication" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6979
Rule of Material Implication/Formulation 1
:$p \implies q \dashv \vdash \neg p \lor q$
{{BeginTableau|\neg p \lor q \vdash p \implies q}} {{Premise|1|\neg p \lor q}} {{Assumption|2|\neg p|Pick the first of the disjuncts ...}} {{Assumption|3|p|Assume its negation ...}} {{NonContradiction|4|2, 3|3|2| ... and demonstrate a contradiction}} {{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can b...
:$p \implies q \dashv \vdash \neg p \lor q$
{{BeginTableau|\neg p \lor q \vdash p \implies q}} {{Premise|1|\neg p \lor q}} {{Assumption|2|\neg p|Pick the first of the disjuncts ...}} {{Assumption|3|p|Assume its negation ...}} {{NonContradiction|4|2, 3|3|2| ... and demonstrate a contradiction}} {{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can b...
Rule of Material Implication/Formulation 1/Reverse Implication/Proof 1
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication/Proof_1
[ "Rule of Material Implication" ]
[]
[]
proofwiki-6980
Rule of Material Implication/Formulation 1
:$p \implies q \dashv \vdash \neg p \lor q$
{{BeginTableau |\neg p \lor q \vdash p \implies q}} {{Premise |1|\neg p \lor q}} {{DeMorgan |2|1|\neg \paren {\neg \neg p \land \neg q}|1|Disjunction}} {{SequentIntro |3|1 |\neg \neg p \implies q |2 |Conditional is Equivalent to Negation of Conjunction with Negative: $\neg \paren {p \land \neg q}...
:$p \implies q \dashv \vdash \neg p \lor q$
{{BeginTableau |\neg p \lor q \vdash p \implies q}} {{Premise |1|\neg p \lor q}} {{DeMorgan |2|1|\neg \paren {\neg \neg p \land \neg q}|1|Disjunction}} {{SequentIntro |3|1 |\neg \neg p \implies q |2 |[[Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implic...
Rule of Material Implication/Formulation 1/Reverse Implication/Proof 2
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication/Proof_2
[ "Rule of Material Implication" ]
[]
[ "Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implication" ]
proofwiki-6981
Rule of Material Implication/Formulation 1/Reverse Implication
:$\neg p \lor q \vdash p \implies q$
{{BeginTableau|\neg p \lor q \vdash p \implies q}} {{Premise|1|\neg p \lor q}} {{Assumption|2|\neg p|Pick the first of the disjuncts ...}} {{Assumption|3|p|Assume its negation ...}} {{NonContradiction|4|2, 3|3|2| ... and demonstrate a contradiction}} {{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can b...
:$\neg p \lor q \vdash p \implies q$
{{BeginTableau|\neg p \lor q \vdash p \implies q}} {{Premise|1|\neg p \lor q}} {{Assumption|2|\neg p|Pick the first of the disjuncts ...}} {{Assumption|3|p|Assume its negation ...}} {{NonContradiction|4|2, 3|3|2| ... and demonstrate a contradiction}} {{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can b...
Rule of Material Implication/Formulation 1/Reverse Implication/Proof 1
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication/Proof_1
[ "Rule of Material Implication" ]
[]
[]
proofwiki-6982
Rule of Material Implication/Formulation 1/Reverse Implication
:$\neg p \lor q \vdash p \implies q$
{{BeginTableau |\neg p \lor q \vdash p \implies q}} {{Premise |1|\neg p \lor q}} {{DeMorgan |2|1|\neg \paren {\neg \neg p \land \neg q}|1|Disjunction}} {{SequentIntro |3|1 |\neg \neg p \implies q |2 |Conditional is Equivalent to Negation of Conjunction with Negative: $\neg \paren {p \land \neg q}...
:$\neg p \lor q \vdash p \implies q$
{{BeginTableau |\neg p \lor q \vdash p \implies q}} {{Premise |1|\neg p \lor q}} {{DeMorgan |2|1|\neg \paren {\neg \neg p \land \neg q}|1|Disjunction}} {{SequentIntro |3|1 |\neg \neg p \implies q |2 |[[Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implic...
Rule of Material Implication/Formulation 1/Reverse Implication/Proof 2
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_1/Reverse_Implication/Proof_2
[ "Rule of Material Implication" ]
[]
[ "Conditional is Equivalent to Negation of Conjunction with Negative/Formulation 1/Forward Implication" ]
proofwiki-6983
Rule of Material Implication/Formulation 2
:$\vdash \paren {p \implies q} \iff \paren {\neg p \lor q}$
{{BeginTableau|\paren {p \implies q} \iff \paren {\neg p \lor q} }} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\neg p \lor q|1|Rule of Material Implication: Formulation 1}} {{Implication|3||\paren {p \implies q} \implies \paren {\neg p \lor q}|1|2}} {{Assumption|4|\neg p \lor q}} {{SequentIntro|5|4|p \implies q|4...
:$\vdash \paren {p \implies q} \iff \paren {\neg p \lor q}$
{{BeginTableau|\paren {p \implies q} \iff \paren {\neg p \lor q} }} {{Assumption|1|p \implies q}} {{SequentIntro|2|1|\neg p \lor q|1|[[Rule of Material Implication/Formulation 1/Forward Implication|Rule of Material Implication: Formulation 1]]}} {{Implication|3||\paren {p \implies q} \implies \paren {\neg p \lor q}|1|2...
Rule of Material Implication/Formulation 2/Proof 1
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Proof_1
[ "Rule of Material Implication" ]
[]
[ "Rule of Material Implication/Formulation 1/Forward Implication", "Rule of Material Implication/Formulation 1/Reverse Implication" ]
proofwiki-6984
Rule of Material Implication/Formulation 2
:$\vdash \paren {p \implies q} \iff \paren {\neg p \lor q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. :<nowiki>$\begin{array}{|ccc|c|cccc|} \hline (p & \implies & q) & \iff & (\neg & p & \lor & q) \\ \hline \F & \T & \F & \T & \T & \F & \T & \F \\ \F & \T & \T & \T & \T...
:$\vdash \paren {p \implies q} \iff \paren {\neg p \lor q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{...
Rule of Material Implication/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2
https://proofwiki.org/wiki/Rule_of_Material_Implication/Formulation_2/Proof_by_Truth_Table
[ "Rule of Material Implication" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6985
Modus Tollendo Ponens/Variant/Formulation 2
:$\vdash \paren {p \lor q} \iff \paren {\neg p \implies q}$
{{BeginTableau|\vdash \paren {p \lor q} \iff \paren {\neg p \implies q} }} {{Assumption|1|p \lor q}} {{SequentIntro|2|1|\neg p \implies q|1|Modus Tollendo Ponens: Formulation 1}} {{Implication|3||\paren {p \lor q} \implies \paren {\neg p \implies q}|1|2}} {{Assumption|4|\neg p \implies q}} {{SequentIntro|5|4|p \lor q|4...
:$\vdash \paren {p \lor q} \iff \paren {\neg p \implies q}$
{{BeginTableau|\vdash \paren {p \lor q} \iff \paren {\neg p \implies q} }} {{Assumption|1|p \lor q}} {{SequentIntro|2|1|\neg p \implies q|1|[[Modus Tollendo Ponens/Variant/Formulation 1/Forward Implication|Modus Tollendo Ponens: Formulation 1]]}} {{Implication|3||\paren {p \lor q} \implies \paren {\neg p \implies q}|1|...
Modus Tollendo Ponens/Variant/Formulation 2/Proof 1
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_2
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_2/Proof_1
[ "Modus Tollendo Ponens" ]
[]
[ "Modus Tollendo Ponens/Variant/Formulation 1/Forward Implication", "Modus Tollendo Ponens/Variant/Formulation 1/Reverse Implication" ]
proofwiki-6986
Modus Tollendo Ponens/Variant/Formulation 2
:$\vdash \paren {p \lor q} \iff \paren {\neg p \implies q}$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations. :<nowiki>$\begin{array}{|ccc|c|cccc|} \hline p & \lor & q & \iff & \neg & p & \implies & q \\ \hline \F & \F & \F & \T & \T & \F & \F & \F \\ \F & \T & \T & \T & \T & \...
:$\vdash \paren {p \lor q} \iff \paren {\neg p \implies q}$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{...
Modus Tollendo Ponens/Variant/Formulation 2/Proof by Truth Table
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_2
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_2/Proof_by_Truth_Table
[ "Modus Tollendo Ponens" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:True", "Definition:Boolean Interpretation" ]
proofwiki-6987
Modus Tollendo Ponens/Variant/Formulation 1
:$p \lor q \dashv \vdash \neg p \implies q$
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. :<nowiki>$\begin{array}{|ccc||cccc|} \hline p & \lor & q & \neg & p & \implies & q \\ \hline \F & \F & \F & \T & \F & \F & \F \\ \F & \T & \T & \T & \F & \T & \T \\ \T & ...
:$p \lor q \dashv \vdash \neg p \implies q$
We apply the [[Method of Truth Tables]]. As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]]. :<nowiki>$\begin{array}{|ccc||cccc|} \hline p ...
Modus Tollendo Ponens/Variant/Formulation 1/Proof by Truth Table
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_1
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_1/Proof_by_Truth_Table
[ "Modus Tollendo Ponens" ]
[]
[ "Method of Truth Tables", "Definition:Truth Value", "Definition:Main Connective/Propositional Logic", "Definition:Boolean Interpretation" ]
proofwiki-6988
Modus Tollendo Ponens/Variant/Formulation 1/Forward Implication
:$p \lor q \vdash \neg p \implies q$
{{BeginTableau|p \lor q \vdash \neg p \implies q}} {{Premise|1|p \lor q}} {{Assumption|2|p|Pick the first of the disjuncts ...}} {{Assumption|3|\neg p|Assume its negation ...}} {{NonContradiction|4|2, 3|2|3|... which introduces a contradiction}} {{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can be der...
:$p \lor q \vdash \neg p \implies q$
{{BeginTableau|p \lor q \vdash \neg p \implies q}} {{Premise|1|p \lor q}} {{Assumption|2|p|Pick the first of the disjuncts ...}} {{Assumption|3|\neg p|Assume its negation ...}} {{NonContradiction|4|2, 3|2|3|... which introduces a contradiction}} {{Explosion|5|2, 3|q|4| ... from a falsehood, ''any'' statement can be der...
Modus Tollendo Ponens/Variant/Formulation 1/Forward Implication
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_1/Forward_Implication
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_1/Forward_Implication
[ "Modus Tollendo Ponens" ]
[]
[ "Category:Modus Tollendo Ponens" ]
proofwiki-6989
Modus Tollendo Ponens/Variant/Formulation 1/Reverse Implication
:$\neg p \implies q \vdash p \lor q$
{{BeginTableau|\neg p \implies q \vdash p \lor q}} {{Premise|1|\neg p \implies q}} {{ExcludedMiddle|2|p \lor \neg p}} {{Assumption|3|p}} {{Addition|4|3|p \lor q|3|1}} {{Assumption|5|\neg p}} {{ModusPonens|6|1, 5|q|1|5}} {{Addition|7|1, 5|p \lor q|6|2}} {{ProofByCases|8|1|p \lor q|2|3|4|5|7}} {{EndTableau|qed}} {{LEM}}
:$\neg p \implies q \vdash p \lor q$
{{BeginTableau|\neg p \implies q \vdash p \lor q}} {{Premise|1|\neg p \implies q}} {{ExcludedMiddle|2|p \lor \neg p}} {{Assumption|3|p}} {{Addition|4|3|p \lor q|3|1}} {{Assumption|5|\neg p}} {{ModusPonens|6|1, 5|q|1|5}} {{Addition|7|1, 5|p \lor q|6|2}} {{ProofByCases|8|1|p \lor q|2|3|4|5|7}} {{EndTableau|qed}} {{LEM}...
Modus Tollendo Ponens/Variant/Formulation 1/Reverse Implication
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_1/Reverse_Implication
https://proofwiki.org/wiki/Modus_Tollendo_Ponens/Variant/Formulation_1/Reverse_Implication
[ "Modus Tollendo Ponens" ]
[]
[]
proofwiki-6990
Radius of Convergence of Power Series over Factorial/Complex Case
Let $\xi \in \C$ be a complex number. Let $\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$. Then $\map f z$ converges absolutely for all $z \in \C$. That is, the radius of convergence of the power series $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {z - \xi}^n} {n!}$ is infinite.
This is a power series in the form $\ds \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$. Applying Radius of Convergence from Limit of Sequence: Complex Case, we find that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } ...
Let $\xi \in \C$ be a [[Definition:Complex Number|complex number]]. Let $\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac {\paren {z - \xi}^n} {n!}$. Then $\map f z$ [[Definition:Absolutely Convergent Series|converges absolutely]] for all $z \in \C$. That is, the [[Definition:Radius of Convergence of Complex Power...
This is a [[Definition:Complex Power Series|power series]] in the form $\ds \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$. Applying [[Radius of Convergence from Limit of Sequence/Complex Case|Radius of Convergence from Limit of Sequence: Complex Case]], we fin...
Radius of Convergence of Power Series over Factorial/Complex Case
https://proofwiki.org/wiki/Radius_of_Convergence_of_Power_Series_over_Factorial/Complex_Case
https://proofwiki.org/wiki/Radius_of_Convergence_of_Power_Series_over_Factorial/Complex_Case
[ "Radius of Convergence of Power Series over Factorial", "Complex Power Series" ]
[ "Definition:Complex Number", "Definition:Absolutely Convergent Series", "Definition:Radius of Convergence/Complex Domain", "Definition:Power Series/Complex Domain" ]
[ "Definition:Power Series/Complex Domain", "Radius of Convergence from Limit of Sequence/Complex Case", "Sequence of Powers of Reciprocals is Null Sequence" ]
proofwiki-6991
Radius of Convergence of Power Series over Factorial/Real Case
Let $\xi \in \R$ be a real number. Let $\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!}$. Then $\map f x$ converges for all $x \in \R$. That is, the interval of convergence of the power series $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!}$ is $\R$.
This is a power series in the form $\ds \sum_{n \mathop= 0}^\infty a_n \paren {x - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$. Applying Radius of Convergence from Limit of Sequence, we find that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \size {\dfrac {a_{n + 1} } {a_n} } | r = \lim_...
Let $\xi \in \R$ be a [[Definition:Real Number|real number]]. Let $\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {\paren {x - \xi}^n} {n!}$. Then $\map f x$ [[Definition:Convergent Function|converges]] for all $x \in \R$. That is, the [[Definition:Interval of Convergence|interval of convergence]] of the [[Defin...
This is a [[Definition:Power Series|power series]] in the form $\ds \sum_{n \mathop= 0}^\infty a_n \paren {x - \xi}^n$ where $\sequence {a_n} = \sequence {\dfrac 1 {n!} }$. Applying [[Radius of Convergence from Limit of Sequence]], we find that: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \size {\dfrac {a_{...
Radius of Convergence of Power Series over Factorial/Real Case
https://proofwiki.org/wiki/Radius_of_Convergence_of_Power_Series_over_Factorial/Real_Case
https://proofwiki.org/wiki/Radius_of_Convergence_of_Power_Series_over_Factorial/Real_Case
[ "Radius of Convergence of Power Series over Factorial", "Real Power Series" ]
[ "Definition:Real Number", "Definition:Convergent Mapping", "Definition:Interval of Convergence", "Definition:Power Series" ]
[ "Definition:Power Series", "Radius of Convergence from Limit of Sequence", "Sequence of Powers of Reciprocals is Null Sequence" ]
proofwiki-6992
Sum of Logarithms/Natural Logarithm
Let $x, y \in \R$ be strictly positive real numbers. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the natural logarithm.
Let $y \in \R_{>0}$ be fixed. Consider the function: :$\map f x = \ln x y - \ln x$ From the definition of the natural logarithm, the Fundamental Theorem of Calculus and the Chain Rule for Derivatives: :$\forall x > 0: \map {f'} x = \dfrac 1 {x y} y - \dfrac 1 x = \dfrac 1 x - \dfrac 1 x = 0$ Thus from Zero Derivative i...
Let $x, y \in \R$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]]. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the [[Definition:Natural Logarithm|natural logarithm]].
Let $y \in \R_{>0}$ be fixed. Consider the [[Definition:Real Function|function]]: :$\map f x = \ln x y - \ln x$ From the definition of the [[Definition:Natural Logarithm|natural logarithm]], the [[Fundamental Theorem of Calculus]] and the [[Chain Rule for Derivatives]]: :$\forall x > 0: \map {f'} x = \dfrac 1 {x y} ...
Sum of Logarithms/Natural Logarithm/Proof 1
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm/Proof_1
[ "Sum of Logarithms" ]
[ "Definition:Strictly Positive/Real Number", "Definition:Natural Logarithm" ]
[ "Definition:Real Function", "Definition:Natural Logarithm", "Fundamental Theorem of Calculus", "Derivative of Composite Function", "Zero Derivative implies Constant Function", "Natural Logarithm of 1 is 0" ]
proofwiki-6993
Sum of Logarithms/Natural Logarithm
Let $x, y \in \R$ be strictly positive real numbers. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the natural logarithm.
{{begin-eqn}} {{eqn | l = \ln x + \ln y | r = \int_1^x \dfrac {\d t} t + \int_1^y \dfrac {\d s} s | c = {{Defof|Natural Logarithm}} }} {{eqn | r = \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t / x} {t / x} | c = Integration by Substitution: $s \mapsto t / x$, $\d s \mapsto \d t / x$, $1 \mapsto...
Let $x, y \in \R$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]]. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the [[Definition:Natural Logarithm|natural logarithm]].
{{begin-eqn}} {{eqn | l = \ln x + \ln y | r = \int_1^x \dfrac {\d t} t + \int_1^y \dfrac {\d s} s | c = {{Defof|Natural Logarithm}} }} {{eqn | r = \int_1^x \dfrac {\d t} t + \int_x^{x y} \dfrac {\d t / x} {t / x} | c = [[Integration by Substitution]]: $s \mapsto t / x$, $\d s \mapsto \d t / x$, $1 \ma...
Sum of Logarithms/Natural Logarithm/Proof 2
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm/Proof_2
[ "Sum of Logarithms" ]
[ "Definition:Strictly Positive/Real Number", "Definition:Natural Logarithm" ]
[ "Integration by Substitution", "Sum of Integrals on Adjacent Intervals for Continuous Functions" ]
proofwiki-6994
Sum of Logarithms/Natural Logarithm
Let $x, y \in \R$ be strictly positive real numbers. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the natural logarithm.
Let $\sequence {f_n}$ be the sequence of mappings $f_n : \R_{>0} \to \R$ defined as: :$\map {f_n} x = n \paren {\sqrt [n] x - 1}$ Let $\map M t = \max \set {\size {t - 1}, \size {\dfrac {t - 1} t} }$ From Bounds of Natural Logarithm: :$\forall t \in \R_{>0}: \size {\map {f_n} t} \le \map M t$ Fix $x, y \in \R_{>0}$. Th...
Let $x, y \in \R$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]]. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the [[Definition:Natural Logarithm|natural logarithm]].
Let $\sequence {f_n}$ be the [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]] $f_n : \R_{>0} \to \R$ defined as: :$\map {f_n} x = n \paren {\sqrt [n] x - 1}$ Let $\map M t = \max \set {\size {t - 1}, \size {\dfrac {t - 1} t} }$ From [[Bounds of Natural Logarithm]]: :$\forall t \in \R_{>0}: \size {\...
Sum of Logarithms/Natural Logarithm/Proof 3
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm/Proof_3
[ "Sum of Logarithms" ]
[ "Definition:Strictly Positive/Real Number", "Definition:Natural Logarithm" ]
[ "Definition:Sequence", "Definition:Mapping", "Bounds of Natural Logarithm", "Absolute Value Function is Completely Multiplicative", "Absolute Value Function is Completely Multiplicative", "Inequality of Product of Unequal Numbers", "Modulus of Limit", "Limit of Bounded Convergent Sequence is Bounded",...
proofwiki-6995
Sum of Logarithms/Natural Logarithm
Let $x, y \in \R$ be strictly positive real numbers. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the natural logarithm.
Recall the definition of the natural logarithm as the definite integral of the reciprocal function: :$\ds \ln x := \int_1^x \frac {\d t} t$ :800px Consider the diagram above. The value of $\ln x$ is represented by the area between the points: :$\tuple {1, 0}, \tuple {1, 1}, \tuple {x, \dfrac 1 x}, \tuple {x, 0}$ which ...
Let $x, y \in \R$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]]. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the [[Definition:Natural Logarithm|natural logarithm]].
Recall the definition of the [[Definition:Natural Logarithm/Positive Real/Definition 1|natural logarithm]] as the [[Definition:Definite Integral|definite integral]] of the [[Definition:Reciprocal|reciprocal function]]: :$\ds \ln x := \int_1^x \frac {\d t} t$ :[[File:SumOfLogarithmsProof4.png|800px]] Consider the d...
Sum of Logarithms/Natural Logarithm/Proof 4
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm/Proof_4
[ "Sum of Logarithms" ]
[ "Definition:Strictly Positive/Real Number", "Definition:Natural Logarithm" ]
[ "Definition:Natural Logarithm/Positive Real/Definition 1", "Definition:Definite Integral", "Definition:Reciprocal", "File:SumOfLogarithmsProof4.png", "Definition:Point", "Definition:Point", "Definition:Point" ]
proofwiki-6996
Sum of Logarithms/Natural Logarithm
Let $x, y \in \R$ be strictly positive real numbers. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the natural logarithm.
{{begin-eqn}} {{eqn | l = x y | r = x \times y | c = }} {{eqn | ll = \leadsto | l = e^{\map \ln {xy} } | r = e^{\ln x} \times e^{\ln y} | c = Exponential of Natural Logarithm }} {{eqn | ll = \leadsto | l = e^{\map \ln {xy} } | r = e^{\ln x + \ln y} | c = Product of Power...
Let $x, y \in \R$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]]. Then: :$\ln x + \ln y = \map \ln {x y}$ where $\ln$ denotes the [[Definition:Natural Logarithm|natural logarithm]].
{{begin-eqn}} {{eqn | l = x y | r = x \times y | c = }} {{eqn | ll = \leadsto | l = e^{\map \ln {xy} } | r = e^{\ln x} \times e^{\ln y} | c = [[Exponential of Natural Logarithm]] }} {{eqn | ll = \leadsto | l = e^{\map \ln {xy} } | r = e^{\ln x + \ln y} | c = [[Product of...
Sum of Logarithms/Natural Logarithm/Proof 5
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm
https://proofwiki.org/wiki/Sum_of_Logarithms/Natural_Logarithm/Proof_5
[ "Sum of Logarithms" ]
[ "Definition:Strictly Positive/Real Number", "Definition:Natural Logarithm" ]
[ "Exponential of Natural Logarithm", "Exponent Combination Laws/Product of Powers" ]
proofwiki-6997
Sum of Logarithms/General Logarithm
Let $x, y, b \in \R$ be strictly positive real numbers such that $b > 1$. Then: :$\log_b x + \log_b y = \map {\log_b} {x y}$ where $\log_b$ denotes the logarithm to base $b$.
{{begin-eqn}} {{eqn | l = \log_b x + \log_b y | r = \map {\log_b} {b^{\log_b x + \log_b y} } | c = {{Defof|General Logarithm}} }} {{eqn | r = \map {\log_b} {\paren {b^{\log_b x} } \paren {b^{\log_b y} } } | c = Product of Powers }} {{eqn | r = \map {\log_b} {x y} | c = {{Defof|General Logarithm}...
Let $x, y, b \in \R$ be [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]] such that $b > 1$. Then: :$\log_b x + \log_b y = \map {\log_b} {x y}$ where $\log_b$ denotes [[Definition:General Logarithm|the logarithm to base $b$]].
{{begin-eqn}} {{eqn | l = \log_b x + \log_b y | r = \map {\log_b} {b^{\log_b x + \log_b y} } | c = {{Defof|General Logarithm}} }} {{eqn | r = \map {\log_b} {\paren {b^{\log_b x} } \paren {b^{\log_b y} } } | c = [[Product of Powers]] }} {{eqn | r = \map {\log_b} {x y} | c = {{Defof|General Logari...
Sum of Logarithms/General Logarithm/Proof 1
https://proofwiki.org/wiki/Sum_of_Logarithms/General_Logarithm
https://proofwiki.org/wiki/Sum_of_Logarithms/General_Logarithm/Proof_1
[ "Sum of Logarithms" ]
[ "Definition:Strictly Positive", "Definition:Real Number", "Definition:General Logarithm" ]
[ "Exponent Combination Laws/Product of Powers" ]
proofwiki-6998
Sum of Logarithms/General Logarithm
Let $x, y, b \in \R$ be strictly positive real numbers such that $b > 1$. Then: :$\log_b x + \log_b y = \map {\log_b} {x y}$ where $\log_b$ denotes the logarithm to base $b$.
{{begin-eqn}} {{eqn | l = \log_b x + \log_b y | r = \frac {\ln x} {\ln b} + \frac {\ln y} {\ln b} | c = Change of Base of Logarithm }} {{eqn | r = \frac {\ln x + \ln y} {\ln b} | c = }} {{eqn | r = \frac {\map \ln {x y} } {\ln b} | c = Sum of Logarithms: Proof for Natural Logarithm }} {{eqn | r...
Let $x, y, b \in \R$ be [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]] such that $b > 1$. Then: :$\log_b x + \log_b y = \map {\log_b} {x y}$ where $\log_b$ denotes [[Definition:General Logarithm|the logarithm to base $b$]].
{{begin-eqn}} {{eqn | l = \log_b x + \log_b y | r = \frac {\ln x} {\ln b} + \frac {\ln y} {\ln b} | c = [[Change of Base of Logarithm]] }} {{eqn | r = \frac {\ln x + \ln y} {\ln b} | c = }} {{eqn | r = \frac {\map \ln {x y} } {\ln b} | c = [[Sum of Logarithms/Natural Logarithm|Sum of Logarithms...
Sum of Logarithms/General Logarithm/Proof 2
https://proofwiki.org/wiki/Sum_of_Logarithms/General_Logarithm
https://proofwiki.org/wiki/Sum_of_Logarithms/General_Logarithm/Proof_2
[ "Sum of Logarithms" ]
[ "Definition:Strictly Positive", "Definition:Real Number", "Definition:General Logarithm" ]
[ "Change of Base of Logarithm", "Sum of Logarithms/Natural Logarithm", "Change of Base of Logarithm" ]
proofwiki-6999
Rule of Simplification/Sequent Form
The '''rule of simplification''' can be symbolised by the sequents:
{{BeginTableau|p \land q \vdash p}} {{Premise|1|p \land q}} {{Simplification|2|1|p|1|1}} {{EndTableau|qed}}
The '''[[Rule of Simplification|rule of simplification]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{BeginTableau|p \land q \vdash p}} {{Premise|1|p \land q}} {{Simplification|2|1|p|1|1}} {{EndTableau|qed}}
Rule of Simplification/Sequent Form/Formulation 1/Form 1/Proof 1
https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form
https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_1/Form_1/Proof_1
[ "Rule of Simplification" ]
[ "Rule of Simplification", "Definition:Sequent" ]
[]