id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-7000 | Rule of Simplification/Sequent Form | The '''rule of simplification''' can be symbolised by the sequents: | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|ccc||c|} \hline
p & \land & q & p \\
\hline
\F & \F & \F & \F \\
\F & \F & \T & \F \\
\T & \F & \F & \T \\
\T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $p \land q$ is true so is $p$.
{{qed}} | The '''[[Rule of Simplification|rule of simplification]]''' can be symbolised by the [[Definition:Sequent|sequents]]: | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|ccc||c|} \hline
p & \land & q & p \\
\hline
\F & \F & \F & \F \\
\F & \F & \T & \F \\
\T & \F & \F & \T \\
\T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $p \land q$ is [[Definition:True|true]] so is $p$.
{{qed}} | Rule of Simplification/Sequent Form/Formulation 1/Form 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_1/Form_1/Proof_by_Truth_Table | [
"Rule of Simplification"
] | [
"Rule of Simplification",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7001 | Rule of Simplification/Sequent Form | The '''rule of simplification''' can be symbolised by the sequents: | {{BeginTableau|p \land q \vdash q}}
{{Premise|1|p \land q}}
{{Simplification|2|1|q|1|2}}
{{EndTableau}}
{{Qed}} | The '''[[Rule of Simplification|rule of simplification]]''' can be symbolised by the [[Definition:Sequent|sequents]]: | {{BeginTableau|p \land q \vdash q}}
{{Premise|1|p \land q}}
{{Simplification|2|1|q|1|2}}
{{EndTableau}}
{{Qed}} | Rule of Simplification/Sequent Form/Formulation 1/Form 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_1/Form_2/Proof_1 | [
"Rule of Simplification"
] | [
"Rule of Simplification",
"Definition:Sequent"
] | [] |
proofwiki-7002 | Rule of Simplification/Sequent Form | The '''rule of simplification''' can be symbolised by the sequents: | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|ccc||c|} \hline
p & \land & q & q \\
\hline
\F & \F & \F & \F \\
\F & \F & \T & \T \\
\T & \F & \F & \F \\
\T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $p \land q$ is true so is $q$.
{{qed}} | The '''[[Rule of Simplification|rule of simplification]]''' can be symbolised by the [[Definition:Sequent|sequents]]: | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|ccc||c|} \hline
p & \land & q & q \\
\hline
\F & \F & \F & \F \\
\F & \F & \T & \T \\
\T & \F & \F & \F \\
\T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $p \land q$ is [[Definition:True|true]] so is $q$.
{{qed}} | Rule of Simplification/Sequent Form/Formulation 1/Form 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_1/Form_2/Proof_by_Truth_Table | [
"Rule of Simplification"
] | [
"Rule of Simplification",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7003 | Rule of Simplification/Sequent Form | The '''rule of simplification''' can be symbolised by the sequents: | We apply the Method of Truth Tables.
:<nowiki>$\begin {array} {|ccc||c|c|} \hline
p & \land & q & p & q \\
\hline
\F & \F & \F & \F & \F \\
\F & \F & \T & \F & \T \\
\T & \F & \F & \T & \F \\
\T & \T & \T & \T & \T \\
\hline
\end {array}$</nowiki>
As can be seen, when $p \land q$ is true so are both $p$ and $q$.
{{qed}... | The '''[[Rule of Simplification|rule of simplification]]''' can be symbolised by the [[Definition:Sequent|sequents]]: | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin {array} {|ccc||c|c|} \hline
p & \land & q & p & q \\
\hline
\F & \F & \F & \F & \F \\
\F & \F & \T & \F & \T \\
\T & \F & \F & \T & \F \\
\T & \T & \T & \T & \T \\
\hline
\end {array}$</nowiki>
As can be seen, when $p \land q$ is [[Definition:True|true]] so ar... | Rule of Simplification/Sequent Form/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_1/Proof_by_Truth_Table | [
"Rule of Simplification"
] | [
"Rule of Simplification",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7004 | Rule of Simplification/Sequent Form | The '''rule of simplification''' can be symbolised by the sequents: | === Form 1 ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1}}
=== Form 2 ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2}} | The '''[[Rule of Simplification|rule of simplification]]''' can be symbolised by the [[Definition:Sequent|sequents]]: | === [[Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1|Form 1]] ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1}}
=== [[Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2|Form 2]] ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2}} | Rule of Simplification/Sequent Form/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_1 | [
"Rule of Simplification"
] | [
"Rule of Simplification",
"Definition:Sequent"
] | [
"Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1",
"Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2"
] |
proofwiki-7005 | Rule of Simplification/Sequent Form | The '''rule of simplification''' can be symbolised by the sequents: | 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>$\begin{array}{|ccc|c|c||c|c|} \hline
p & \land & q & p & q & p \land q \implies p & p \land q \implies q \\
\hline
\F & \F & \F & \F & \F & \T & \T \\
\F & \F... | The '''[[Rule of Simplification|rule of simplification]]''' can be symbolised by the [[Definition:Sequent|sequents]]: | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{... | Rule of Simplification/Sequent Form/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_by_Truth_Table | [
"Rule of Simplification"
] | [
"Rule of Simplification",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-7006 | Rule of Conjunction/Sequent Form | The Rule of Conjunction can be symbolised in sequent form as follows: | {{BeginTableau|p, q \vdash p \land q}}
{{Premise|1|p}}
{{Premise|2|q}}
{{Conjunction|3|1, 2|p \land q|1|2}}
{{EndTableau|qed}} | The [[Rule of Conjunction]] can be symbolised in [[Definition:Sequent|sequent]] form as follows: | {{BeginTableau|p, q \vdash p \land q}}
{{Premise|1|p}}
{{Premise|2|q}}
{{Conjunction|3|1, 2|p \land q|1|2}}
{{EndTableau|qed}} | Rule of Conjunction/Sequent Form/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Conjunction/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Conjunction/Sequent_Form/Formulation_1/Proof_1 | [
"Rule of Conjunction"
] | [
"Rule of Conjunction",
"Definition:Sequent"
] | [] |
proofwiki-7007 | Rule of Conjunction/Sequent Form | The Rule of Conjunction can be symbolised in sequent form as follows: | We apply the Method of Truth Tables.
$\begin{array}{|c|c||ccc|} \hline
p & q & p & \land & q\\
\hline
\F & \F & \F & \F & \F \\
\F & \T & \F & \F & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, only when both $p$ and $q$ are true, then so is $p \land q$.
{{qed}} | The [[Rule of Conjunction]] can be symbolised in [[Definition:Sequent|sequent]] form as follows: | We apply the [[Method of Truth Tables]].
$\begin{array}{|c|c||ccc|} \hline
p & q & p & \land & q\\
\hline
\F & \F & \F & \F & \F \\
\F & \T & \F & \F & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, only when both $p$ and $q$ are [[Definition:True|true]], then so is $p \... | Rule of Conjunction/Sequent Form/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Conjunction/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Conjunction/Sequent_Form/Formulation_1/Proof_by_Truth_Table | [
"Rule of Conjunction"
] | [
"Rule of Conjunction",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7008 | Radius of Convergence of Derivative of Complex Power Series | Let $\xi \in \C$.
For all $z \in \C$, define the power series:
:$\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$
and:
:$\ds \map {S'} z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
Let $R$ be the radius of convergence of $\map S z$, and let $R'$ be the radius of convergence of $\map ... | Suppose that $z \in \C$ with $\cmod {z - \xi} < R'$.
Then $\map {S'} z$ converges absolutely by Existence of Radius of Convergence of Complex Power Series, so:
{{begin-eqn}}
{{eqn | l = 1
| o = \ge
| r = \limsup_{n \mathop \to \infty} \cmod {n a_n \paren {z - \xi}^{n - 1} }^{1 / n}
| c = $n$th Root Te... | Let $\xi \in \C$.
For all $z \in \C$, define the [[Definition:Complex Power Series|power series]]:
:$\ds \map S z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$
and:
:$\ds \map {S'} z = \sum_{n \mathop = 1}^\infty n a_n \paren {z - \xi}^{n - 1}$
Let $R$ be the [[Definition:Radius of Convergence of Complex Powe... | Suppose that $z \in \C$ with $\cmod {z - \xi} < R'$.
Then $\map {S'} z$ [[Definition:Absolutely Convergent Series|converges absolutely]] by [[Existence of Radius of Convergence of Complex Power Series]], so:
{{begin-eqn}}
{{eqn | l = 1
| o = \ge
| r = \limsup_{n \mathop \to \infty} \cmod {n a_n \paren {z ... | Radius of Convergence of Derivative of Complex Power Series | https://proofwiki.org/wiki/Radius_of_Convergence_of_Derivative_of_Complex_Power_Series | https://proofwiki.org/wiki/Radius_of_Convergence_of_Derivative_of_Complex_Power_Series | [
"Complex Power Series",
"Complex Analysis"
] | [
"Definition:Power Series/Complex Domain",
"Definition:Radius of Convergence/Complex Domain"
] | [
"Definition:Absolutely Convergent Series",
"Existence of Radius of Convergence of Complex Power Series",
"Nth Root Test",
"Nth Root Test",
"Definition:Absolutely Convergent Series",
"Existence of Radius of Convergence of Complex Power Series",
"Definition:Absolutely Convergent Series",
"Nth Root Test"... |
proofwiki-7009 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | {{BeginTableau|q \vdash p \lor q}}
{{Premise|1|q}}
{{Addition|2|1|p \lor q|1|2}}
{{EndTableau|qed}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | {{BeginTableau|q \vdash p \lor q}}
{{Premise|1|q}}
{{Addition|2|1|p \lor q|1|2}}
{{EndTableau|qed}} | Rule of Addition/Sequent Form/Form 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Form_2/Proof_1 | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [] |
proofwiki-7010 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | {{BeginTableau|p \vdash p \lor q}}
{{Premise|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{EndTableau|qed}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | {{BeginTableau|p \vdash p \lor q}}
{{Premise|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{EndTableau|qed}} | Rule of Addition/Sequent Form/Formulation 1/Form 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_1/Form_1/Proof_1 | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [] |
proofwiki-7011 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | We apply the Method of Truth Tables.
$\begin{array}{|c||ccc|} \hline
p & p & \lor & q \\
\hline
\F & \F & \F & \F \\
\F & \F & \T & \T \\
\T & \T & \T & \F \\
\T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, when $p$ is true so is $p \lor q$.
{{qed}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | We apply the [[Method of Truth Tables]].
$\begin{array}{|c||ccc|} \hline
p & p & \lor & q \\
\hline
\F & \F & \F & \F \\
\F & \F & \T & \T \\
\T & \T & \T & \F \\
\T & \T & \T & \T \\
\hline
\end{array}$
As can be seen, when $p$ is [[Definition:True|true]] so is $p \lor q$.
{{qed}} | Rule of Addition/Sequent Form/Formulation 1/Form 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_1/Form_1/Proof_by_Truth_Table | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7012 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|c||ccc|} \hline
q & p & \lor & q \\
\hline
\F & \F & \F & \F \\
\T & \F & \T & \T \\
\F & \T & \T & \F \\
\T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $q$ is true so is $p \lor q$.
{{qed}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|c||ccc|} \hline
q & p & \lor & q \\
\hline
\F & \F & \F & \F \\
\T & \F & \T & \T \\
\F & \T & \T & \F \\
\T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $q$ is [[Definition:True|true]] so is $p \lor q$.
{{qed}} | Rule of Addition/Sequent Form/Formulation 1/Form 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_1/Form_2/Proof_by_Truth_Table | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7013 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | {{BeginTableau|p \implies \paren {p \lor q} }}
{{Premise|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{Implication|3||p \implies \paren {p \lor q}|1|3}}
{{EndTableau}}
{{Qed}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | {{BeginTableau|p \implies \paren {p \lor q} }}
{{Premise|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{Implication|3||p \implies \paren {p \lor q}|1|3}}
{{EndTableau}}
{{Qed}} | Rule of Addition/Sequent Form/Formulation 2/Form 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_2/Form_1/Proof_1 | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [] |
proofwiki-7014 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | {{BeginTableau|p \implies \paren {p \lor q}|Instance 2 of the Hilbert-style systems}}
{{TableauLine
| n = 1
| f = q \implies \paren {p \lor q}
| rlnk = Definition:Hilbert Proof System/Instance 2
| rtxt = Axiom $\text A 2$
}}
{{TableauLine
| n = 2
| f = p \implies \paren {q \lor p}
| rlnk = Definition:Hilbert Pro... | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | {{BeginTableau|p \implies \paren {p \lor q}|[[Definition:Hilbert Proof System/Instance 2|Instance 2 of the Hilbert-style systems]]}}
{{TableauLine
| n = 1
| f = q \implies \paren {p \lor q}
| rlnk = Definition:Hilbert Proof System/Instance 2
| rtxt = Axiom $\text A 2$
}}
{{TableauLine
| n = 2
| f = p \implies \pa... | Rule of Addition/Sequent Form/Formulation 2/Form 1/Proof 2 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_2/Form_1/Proof_2 | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Definition:Hilbert Proof System/Instance 2"
] |
proofwiki-7015 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | {{BeginTableau|q \implies \paren {p \lor q} }}
{{Premise|1|q}}
{{Addition|2|1|p \lor q|1|2}}
{{Implication|3||q \implies \paren {p \lor q}|1|3}}
{{EndTableau}}
{{Qed}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | {{BeginTableau|q \implies \paren {p \lor q} }}
{{Premise|1|q}}
{{Addition|2|1|p \lor q|1|2}}
{{Implication|3||q \implies \paren {p \lor q}|1|3}}
{{EndTableau}}
{{Qed}} | Rule of Addition/Sequent Form/Formulation 2/Form 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_2/Form_2/Proof_1 | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [] |
proofwiki-7016 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | === Form 1 ===
{{:Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 1}}
=== Form 2 ===
{{:Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 2}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | === [[Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 1|Form 1]] ===
{{:Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 1}}
=== [[Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 2|Form 2]] ===
{{:Rule of Addition/Sequent Form/Formulation 2/Proof 1/Form 2}} | Rule of Addition/Sequent Form/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_2/Proof_1 | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Rule of Addition/Sequent Form/Formulation 2/Form 1/Proof 1",
"Rule of Addition/Sequent Form/Formulation 2/Form 2/Proof 1"
] |
proofwiki-7017 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives are $T$ for all boolean interpretations.
:<nowiki>$\begin{array}{|c|c|ccccc|ccccc|} \hline
p & q & p & \implies & (p & \lor & q) & q & \implies & (p & \lor & q) \\
\hline
\F & \F & \F & \T & \F & \F & \F & \F ... | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | 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]] are $T$ for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|c|c|ccccc|ccccc|} \... | Rule of Addition/Sequent Form/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Formulation_2/Proof_by_Truth_Table | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7018 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | === Form 1 ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 1}}
=== Form 2 ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 2}} | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | === [[Rule of Addition/Sequent Form/Proof 1/Form 1|Form 1]] ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 1}}
=== [[Rule of Addition/Sequent Form/Proof 1/Form 2|Form 2]] ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 2}} | Rule of Addition/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Proof_1 | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Rule of Addition/Sequent Form/Formulation 1/Form 1/Proof 1",
"Rule of Addition/Sequent Form/Form 2/Proof 1"
] |
proofwiki-7019 | Rule of Addition/Sequent Form | The '''rule of addition''' can be symbolised by the sequents:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}} | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|c|c||ccc|} \hline
p & q & p & \lor & q\\
\hline
\F & \F & \F & \F & \F \\
\F & \T & \F & \T & \T \\
\T & \F & \T & \T & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, whenever either $p$ or $q$ (or both) are true, then so is $p ... | The '''[[Rule of Addition|rule of addition]]''' can be symbolised by the [[Definition:Sequent|sequents]]:
{{begin-eqn}}
{{eqn | n = 1
| l = p
| o =
}}
{{eqn | ll= \vdash
| l = p \lor q
| o =
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | n = 2
| l = q
| o =
}}
{{eqn | ll= \vdash
|... | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|c|c||ccc|} \hline
p & q & p & \lor & q\\
\hline
\F & \F & \F & \F & \F \\
\F & \T & \F & \T & \T \\
\T & \F & \T & \T & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, whenever either $p$ or $q$ (or both) are [[Definition:T... | Rule of Addition/Sequent Form/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Proof_by_Truth_Table | [
"Rule of Addition",
"Disjunction"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7020 | Rule of Addition/Sequent Form/Proof 1 | The Rule of Addition can be symbolised by the sequents:
:$(1): \quad p \vdash p \lor q$
:$(2): \quad q \vdash p \lor q$ | === Form 1 ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 1}}
=== Form 2 ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 2}} | The [[Rule of Addition]] can be symbolised by the [[Definition:Sequent|sequents]]:
:$(1): \quad p \vdash p \lor q$
:$(2): \quad q \vdash p \lor q$ | === [[Rule of Addition/Sequent Form/Proof 1/Form 1|Form 1]] ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 1}}
=== [[Rule of Addition/Sequent Form/Proof 1/Form 2|Form 2]] ===
{{:Rule of Addition/Sequent Form/Proof 1/Form 2}} | Rule of Addition/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Proof_1 | https://proofwiki.org/wiki/Rule_of_Addition/Sequent_Form/Proof_1 | [
"Rule of Addition"
] | [
"Rule of Addition",
"Definition:Sequent"
] | [
"Rule of Addition/Sequent Form/Formulation 1/Form 1/Proof 1",
"Rule of Addition/Sequent Form/Form 2/Proof 1"
] |
proofwiki-7021 | Proof by Cases/Sequent Form | Proof by Cases can be symbolised by the sequent:
:$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$ | {{BeginTableau|p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r}}
{{Premise | 1 | p \lor q}}
{{Assumption | 2 | p}}
{{TableauLine | n = 3
| pool = 2
| f = r
| rlnk = Definition:By Hypothesis
| rtxt = By hypothesis
| dep = 2
... | [[Proof by Cases]] can be symbolised by the [[Definition:Sequent|sequent]]:
:$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$ | {{BeginTableau|p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r}}
{{Premise | 1 | p \lor q}}
{{Assumption | 2 | p}}
{{TableauLine | n = 3
| pool = 2
| f = r
| rlnk = Definition:By Hypothesis
| rtxt = By hypothesis
| dep = 2
... | Proof by Cases/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Proof_by_Cases/Sequent_Form | https://proofwiki.org/wiki/Proof_by_Cases/Sequent_Form/Proof_1 | [
"Proof by Cases"
] | [
"Proof by Cases",
"Definition:Sequent"
] | [] |
proofwiki-7022 | Proof by Cases/Sequent Form | Proof by Cases can be symbolised by the sequent:
:$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$ | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|ccc|ccc|ccc||c|} \hline
p & \lor & q & p & \implies & r & q & \implies & r & r \\
\hline
\F & \F & \F & \F & \T & \F & \F & \T & \F & \F \\
\F & \F & \F & \F & \T & \T & \F & \T & \T & \T \\
\F & \T & \T & \F & \T & \F & \T & \F & \F & \F \\
\F & \T & \T & \... | [[Proof by Cases]] can be symbolised by the [[Definition:Sequent|sequent]]:
:$p \lor q, \paren {p \vdash r}, \paren {q \vdash r} \vdash r$ | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|ccc|ccc|ccc||c|} \hline
p & \lor & q & p & \implies & r & q & \implies & r & r \\
\hline
\F & \F & \F & \F & \T & \F & \F & \T & \F & \F \\
\F & \F & \F & \F & \T & \T & \F & \T & \T & \T \\
\F & \T & \T & \F & \T & \F & \T & \F & \F & \F \\
\F & \T & \... | Proof by Cases/Sequent Form/Proof by Truth Table | https://proofwiki.org/wiki/Proof_by_Cases/Sequent_Form | https://proofwiki.org/wiki/Proof_by_Cases/Sequent_Form/Proof_by_Truth_Table | [
"Proof by Cases"
] | [
"Proof by Cases",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True"
] |
proofwiki-7023 | Modus Ponendo Ponens/Sequent Form | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = p
| o =
}}
{{eqn | ll= \vdash
| l = q
| o =
}}
{{end-eqn}} | {{BeginTableau|p \implies q, p \vdash q}}
{{Premise|1|p \implies q}}
{{Premise|2|p}}
{{ModusPonens|3|1, 2|q|1|2}}
{{EndTableau|qed}} | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = p
| o =
}}
{{eqn | ll= \vdash
| l = q
| o =
}}
{{end-eqn}} | {{BeginTableau|p \implies q, p \vdash q}}
{{Premise|1|p \implies q}}
{{Premise|2|p}}
{{ModusPonens|3|1, 2|q|1|2}}
{{EndTableau|qed}} | Modus Ponendo Ponens/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Sequent_Form | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Sequent_Form/Proof_1 | [
"Modus Ponendo Ponens"
] | [] | [] |
proofwiki-7024 | Modus Ponendo Ponens/Sequent Form | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = p
| o =
}}
{{eqn | ll= \vdash
| l = q
| o =
}}
{{end-eqn}} | {{BeginTableau|p \implies q, p \vdash q}}
{{Premise|1|p \implies q}}
{{Premise|2|p}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q|1|3}}
{{Assumption|5|\lnot p}}
{{NonContradiction|6|2, 5|2|5}}
{{Explosion|7|2, 5|q|6}}
{{ExcludedMiddle|8|p \lor \lnot p}}
{{ProofByCases|9|1, 2|q|8|3|4|5|7}}
{{EndTableau|qed}} | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = p
| o =
}}
{{eqn | ll= \vdash
| l = q
| o =
}}
{{end-eqn}} | {{BeginTableau|p \implies q, p \vdash q}}
{{Premise|1|p \implies q}}
{{Premise|2|p}}
{{Assumption|3|p}}
{{ModusPonens|4|1, 3|q|1|3}}
{{Assumption|5|\lnot p}}
{{NonContradiction|6|2, 5|2|5}}
{{Explosion|7|2, 5|q|6}}
{{ExcludedMiddle|8|p \lor \lnot p}}
{{ProofByCases|9|1, 2|q|8|3|4|5|7}}
{{EndTableau|qed}} | Modus Ponendo Ponens/Sequent Form/Proof 2 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Sequent_Form | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Sequent_Form/Proof_2 | [
"Modus Ponendo Ponens"
] | [] | [] |
proofwiki-7025 | Modus Ponendo Ponens/Sequent Form | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = p
| o =
}}
{{eqn | ll= \vdash
| l = q
| o =
}}
{{end-eqn}} | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|c|ccc||c|} \hline
p & p & \implies & q & q\\
\hline
\F & \F & \T & \F & \F \\
\F & \F & \T & \T & \T \\
\T & \T & \F & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $p$ is true, and so is $p \implies q$, then $q$ is a... | {{begin-eqn}}
{{eqn | l = p
| o = \implies
| r = q
}}
{{eqn | l = p
| o =
}}
{{eqn | ll= \vdash
| l = q
| o =
}}
{{end-eqn}} | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|c|ccc||c|} \hline
p & p & \implies & q & q\\
\hline
\F & \F & \T & \F & \F \\
\F & \F & \T & \T & \T \\
\T & \T & \F & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen, when $p$ is [[Definition:True|true]], and so is $p ... | Modus Ponendo Ponens/Sequent Form/Proof by Truth Table | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Sequent_Form | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Sequent_Form/Proof_by_Truth_Table | [
"Modus Ponendo Ponens"
] | [] | [
"Method of Truth Tables",
"Definition:True",
"Definition:True"
] |
proofwiki-7026 | Logarithm of Power/Natural Logarithm | Let $x \in \R$ be a strictly positive real number.
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \R$ be any real number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\map \ln {x^r} = r \ln x$ | Consider the function $\map f x = \map \ln {x^r} - r \ln x$.
Then from:
:The definition of the natural logarithm
:The Fundamental Theorem of Calculus
:The Power Rule for Derivatives
:The Chain Rule for Derivatives:
:$\forall x > 0: \map {f'} x = \dfrac 1 {x^r} r x^{r-1} - \dfrac r x = 0$
Thus from Zero Derivative impli... | Let $x \in \R$ be a [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real number]].
Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
Let $r \in \R$ be any [[Definition:Real Number|real number]].
Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarith... | Consider the [[Definition:Real Function|function]] $\map f x = \map \ln {x^r} - r \ln x$.
Then from:
:The definition of the [[Definition:Natural Logarithm|natural logarithm]]
:The [[Fundamental Theorem of Calculus]]
:The [[Power Rule for Derivatives]]
:The [[Chain Rule for Derivatives]]:
:$\forall x > 0: \map {f'} x ... | Logarithm of Power/Natural Logarithm/Proof 1 | https://proofwiki.org/wiki/Logarithm_of_Power/Natural_Logarithm | https://proofwiki.org/wiki/Logarithm_of_Power/Natural_Logarithm/Proof_1 | [
"Logarithm of Power"
] | [
"Definition:Strictly Positive",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Natural Logarithm"
] | [
"Definition:Real Function",
"Definition:Natural Logarithm",
"Fundamental Theorem of Calculus",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Zero Derivative implies Constant Function",
"Natural Logarithm of 1 is 0"
] |
proofwiki-7027 | Logarithm of Power/Natural Logarithm | Let $x \in \R$ be a strictly positive real number.
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \R$ be any real number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\map \ln {x^r} = r \ln x$ | {{begin-eqn}}
{{eqn | l = \ln a
| r = b
}}
{{eqn | ll= \leadstoandfrom
| l = e^b
| r = a
}}
{{eqn | ll= \leadsto
| l = \paren {e^b}^c
| r = a^c
}}
{{eqn | ll= \leadsto
| l = e^{c b}
| r = a^c
| c = Exponential of Product
}}
{{eqn | ll= \leadsto
| l = \ln e^{c b}
... | Let $x \in \R$ be a [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real number]].
Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
Let $r \in \R$ be any [[Definition:Real Number|real number]].
Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarith... | {{begin-eqn}}
{{eqn | l = \ln a
| r = b
}}
{{eqn | ll= \leadstoandfrom
| l = e^b
| r = a
}}
{{eqn | ll= \leadsto
| l = \paren {e^b}^c
| r = a^c
}}
{{eqn | ll= \leadsto
| l = e^{c b}
| r = a^c
| c = [[Exponential of Product]]
}}
{{eqn | ll= \leadsto
| l = \ln e^{c b}... | Logarithm of Power/Natural Logarithm/Proof 2 | https://proofwiki.org/wiki/Logarithm_of_Power/Natural_Logarithm | https://proofwiki.org/wiki/Logarithm_of_Power/Natural_Logarithm/Proof_2 | [
"Logarithm of Power"
] | [
"Definition:Strictly Positive",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Natural Logarithm"
] | [
"Exponential of Product",
"Exponential of Natural Logarithm",
"Definition:By Hypothesis"
] |
proofwiki-7028 | Logarithm of Power/Natural Logarithm | Let $x \in \R$ be a strictly positive real number.
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \R$ be any real number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
:$\map \ln {x^r} = r \ln x$ | Here we adopt the definition of $\ln x$ to be:
:$\ds \ln x := \int_1^x \dfrac {\d t} t$
{{begin-eqn}}
{{eqn | l = \map \ln {x^r}
| r = \int_1^{x^r} \dfrac {\d t} t
| c = {{Defof|Natural Logarithm}}
}}
{{eqn | r = \int_1^x \dfrac {r t^{r - 1} \rd t} {t^r}
| c = Integration by Substitution: $t \mapsto t... | Let $x \in \R$ be a [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real number]].
Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
Let $r \in \R$ be any [[Definition:Real Number|real number]].
Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarith... | Here we adopt the definition of $\ln x$ to be:
:$\ds \ln x := \int_1^x \dfrac {\d t} t$
{{begin-eqn}}
{{eqn | l = \map \ln {x^r}
| r = \int_1^{x^r} \dfrac {\d t} t
| c = {{Defof|Natural Logarithm}}
}}
{{eqn | r = \int_1^x \dfrac {r t^{r - 1} \rd t} {t^r}
| c = [[Integration by Substitution]]: $t \map... | Logarithm of Power/Natural Logarithm/Proof 3 | https://proofwiki.org/wiki/Logarithm_of_Power/Natural_Logarithm | https://proofwiki.org/wiki/Logarithm_of_Power/Natural_Logarithm/Proof_3 | [
"Logarithm of Power"
] | [
"Definition:Strictly Positive",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Natural Logarithm"
] | [
"Integration by Substitution",
"Primitive of Constant Multiple of Function"
] |
proofwiki-7029 | Logarithm of Power/General Logarithm | Let $x \in \R$ be a strictly positive real number.
Let $a \in \R$ be a real number such that $a > 1$.
Let $r \in \R$ be any real number.
Let $\log_a x$ be the logarithm to the base $a$ of $x$.
Then:
:$\map {\log_a} {x^r} = r \log_a x$ | Let $y = r \log_a x$.
Then:
{{begin-eqn}}
{{eqn | l = a^y
| r = a^{r \log_a x}
| c =
}}
{{eqn | r = \paren {a^{\log_a x} }^r
| c = Exponent Combination Laws
}}
{{eqn | r = x^r
| c = {{Defof|General Logarithm|Logarithm base $a$}}
}}
{{end-eqn}}
The result follows by taking logs base $a$ of both ... | Let $x \in \R$ be a [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real number]].
Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
Let $r \in \R$ be any [[Definition:Real Number|real number]].
Let $\log_a x$ be the [[Definition:General Logarithm|logarithm to ... | Let $y = r \log_a x$.
Then:
{{begin-eqn}}
{{eqn | l = a^y
| r = a^{r \log_a x}
| c =
}}
{{eqn | r = \paren {a^{\log_a x} }^r
| c = [[Exponent Combination Laws]]
}}
{{eqn | r = x^r
| c = {{Defof|General Logarithm|Logarithm base $a$}}
}}
{{end-eqn}}
The result follows by [[Definition:General L... | Logarithm of Power/General Logarithm | https://proofwiki.org/wiki/Logarithm_of_Power/General_Logarithm | https://proofwiki.org/wiki/Logarithm_of_Power/General_Logarithm | [
"Logarithm of Power"
] | [
"Definition:Strictly Positive",
"Definition:Real Number",
"Definition:Real Number",
"Definition:Real Number",
"Definition:General Logarithm"
] | [
"Exponent Combination Laws",
"Definition:General Logarithm"
] |
proofwiki-7030 | Power Set with Union and Intersection forms Boolean Algebra | Let $S$ be a set, and let $\powerset S$ be its power set.
Denote with $\cup$, $\cap$ and $\complement$ the operations of union, intersection and complement on $\powerset S$, respectively.
Then $\struct {\powerset S, \cup, \cap, \complement}$ is a Boolean algebra. | Taking the criteria for definition 1 of a Boolean algebra in turn: | Let $S$ be a [[Definition:Set|set]], and let $\powerset S$ be its [[Definition:Power Set|power set]].
Denote with $\cup$, $\cap$ and $\complement$ the operations of [[Definition:Set Union|union]], [[Definition:Set Intersection|intersection]] and [[Definition:Complement|complement]] on $\powerset S$, respectively.
Th... | Taking the criteria for [[Definition:Boolean Algebra/Definition 1|definition 1 of a Boolean algebra]] in turn: | Power Set with Union and Intersection forms Boolean Algebra | https://proofwiki.org/wiki/Power_Set_with_Union_and_Intersection_forms_Boolean_Algebra | https://proofwiki.org/wiki/Power_Set_with_Union_and_Intersection_forms_Boolean_Algebra | [
"Power Set",
"Boolean Algebras"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Set Union",
"Definition:Set Intersection",
"Definition:Complement",
"Definition:Boolean Algebra"
] | [
"Definition:Boolean Algebra/Definition 1",
"Definition:Boolean Algebra/Definition 1"
] |
proofwiki-7031 | Rule of Implication/Sequent Form | The '''rule of implication''' can be symbolised by the sequent:
{{begin-eqn}}
{{eqn | l = \paren {p \vdash q}
| o =
}}
{{eqn | ll= \vdash
| l = p \implies q
| o =
}}
{{end-eqn}} | * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Rule of Simplification/Sequent Form/Formulation 1/Form 2|next = Indirect Proof}}: $3$: The Method of Deduction: $3.5$: The Rule of Conditional Proof
Category:Rule of Implication
sok3mffh8m5eu15dn2enblza3f6myp6 | The '''[[Rule of Implication|rule of implication]]''' can be symbolised by the [[Definition:Sequent|sequent]]:
{{begin-eqn}}
{{eqn | l = \paren {p \vdash q}
| o =
}}
{{eqn | ll= \vdash
| l = p \implies q
| o =
}}
{{end-eqn}} | * {{BookReference|Symbolic Logic|1973|Irving M. Copi|ed = 4th|edpage = Fourth Edition|prev = Rule of Simplification/Sequent Form/Formulation 1/Form 2|next = Indirect Proof}}: $3$: The Method of Deduction: $3.5$: The Rule of Conditional Proof
[[Category:Rule of Implication]]
sok3mffh8m5eu15dn2enblza3f6myp6 | Rule of Implication/Sequent Form | https://proofwiki.org/wiki/Rule_of_Implication/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Implication/Sequent_Form | [
"Rule of Implication"
] | [
"Rule of Implication",
"Definition:Sequent"
] | [
"Category:Rule of Implication"
] |
proofwiki-7032 | Rule of Implication/Sequent Form | The '''rule of implication''' can be symbolised by the sequent:
{{begin-eqn}}
{{eqn | l = \paren {p \vdash q}
| o =
}}
{{eqn | ll= \vdash
| l = p \implies q
| o =
}}
{{end-eqn}} | {{BeginTableau|\paren {p \vdash q} \vdash p \implies q}}
{{Premise|1|p}}
{{TableauLine | n = 2
| pool = 1
| f = q
| rlnk = Definition:By Hypothesis
| rtxt = By hypothesis
| dep = 1
| c = as $p \vdash q$
}}
{{Implication|3|1|p ... | The '''[[Rule of Implication|rule of implication]]''' can be symbolised by the [[Definition:Sequent|sequent]]:
{{begin-eqn}}
{{eqn | l = \paren {p \vdash q}
| o =
}}
{{eqn | ll= \vdash
| l = p \implies q
| o =
}}
{{end-eqn}} | {{BeginTableau|\paren {p \vdash q} \vdash p \implies q}}
{{Premise|1|p}}
{{TableauLine | n = 2
| pool = 1
| f = q
| rlnk = Definition:By Hypothesis
| rtxt = By hypothesis
| dep = 1
| c = as $p \vdash q$
}}
{{Implication|3|1|p ... | Rule of Implication/Sequent Form/Proof 1 | https://proofwiki.org/wiki/Rule_of_Implication/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Implication/Sequent_Form/Proof_1 | [
"Rule of Implication"
] | [
"Rule of Implication",
"Definition:Sequent"
] | [] |
proofwiki-7033 | Rule of Implication/Sequent Form | The '''rule of implication''' can be symbolised by the sequent:
{{begin-eqn}}
{{eqn | l = \paren {p \vdash q}
| o =
}}
{{eqn | ll= \vdash
| l = p \implies q
| o =
}}
{{end-eqn}} | We apply the Method of Truth Tables.
:<nowiki>$\begin{array}{|c|c||ccc|} \hline
p & q & p & \implies & q\\
\hline
\F & \F & \F & \T & \F \\
\F & \T & \F & \T & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen by inspection, only when $p$ is true and $q$ is false, the... | The '''[[Rule of Implication|rule of implication]]''' can be symbolised by the [[Definition:Sequent|sequent]]:
{{begin-eqn}}
{{eqn | l = \paren {p \vdash q}
| o =
}}
{{eqn | ll= \vdash
| l = p \implies q
| o =
}}
{{end-eqn}} | We apply the [[Method of Truth Tables]].
:<nowiki>$\begin{array}{|c|c||ccc|} \hline
p & q & p & \implies & q\\
\hline
\F & \F & \F & \T & \F \\
\F & \T & \F & \T & \T \\
\T & \F & \T & \F & \F \\
\T & \T & \T & \T & \T \\
\hline
\end{array}$</nowiki>
As can be seen by inspection, only when $p$ is [[Definition:True|tr... | Rule of Implication/Sequent Form/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Implication/Sequent_Form | https://proofwiki.org/wiki/Rule_of_Implication/Sequent_Form/Proof_by_Truth_Table | [
"Rule of Implication"
] | [
"Rule of Implication",
"Definition:Sequent"
] | [
"Method of Truth Tables",
"Definition:True",
"Definition:False",
"Definition:False"
] |
proofwiki-7034 | Sum of Absolutely Convergent Series | Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n}$ is absolutely convergent, and:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n + b_n} = \sum_{n \mathop = 1... | Let $\epsilon \in \R_{>0}$.
From Tail of Convergent Series tends to Zero, it follows that there exists $M \in \N$ such that:
:$\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$
and:
:$\ds\sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$
For all $m \ge M$, it follows that:
{{begin-eqn}}... | Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two [[Definition:Real Number|real]] or [[Definition:Complex Number|complex]] [[Definition:Series|series]] that are [[Definition:Absolutely Convergent Series|absolutely convergent]].
Then the series $\ds \sum_{n \mathop = 1}^\infty ... | Let $\epsilon \in \R_{>0}$.
From [[Tail of Convergent Series tends to Zero]], it follows that there exists $M \in \N$ such that:
:$\ds \sum_{n \mathop = M + 1}^\infty \cmod {a_n} < \dfrac \epsilon 2$
and:
:$\ds\sum_{n \mathop = M + 1}^\infty \cmod {b_n} < \dfrac \epsilon 2$
For all $m \ge M$, it follows that:
{{begi... | Sum of Absolutely Convergent Series | https://proofwiki.org/wiki/Sum_of_Absolutely_Convergent_Series | https://proofwiki.org/wiki/Sum_of_Absolutely_Convergent_Series | [
"Absolute Convergence"
] | [
"Definition:Real Number",
"Definition:Complex Number",
"Definition:Series",
"Definition:Absolutely Convergent Series",
"Definition:Absolutely Convergent Series"
] | [
"Tail of Convergent Series tends to Zero",
"Triangle Inequality",
"Definition:Convergent Series",
"Definition:Absolutely Convergent Series",
"Triangle Inequality"
] |
proofwiki-7035 | Rule of Simplification/Sequent Form/Formulation 2 | {{begin-eqn}}
{{eqn | n = 1
| l = \vdash p \land q
| o = \implies
| r = p
}}
{{eqn | n = 2
| l = \vdash p \land q
| o = \implies
| r = q
}}
{{end-eqn}} | === Form 1 ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1}}
=== Form 2 ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2}} | {{begin-eqn}}
{{eqn | n = 1
| l = \vdash p \land q
| o = \implies
| r = p
}}
{{eqn | n = 2
| l = \vdash p \land q
| o = \implies
| r = q
}}
{{end-eqn}} | === [[Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1|Form 1]] ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1}}
=== [[Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2|Form 2]] ===
{{:Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2}} | Rule of Simplification/Sequent Form/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_1 | [
"Rule of Simplification"
] | [] | [
"Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1",
"Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2"
] |
proofwiki-7036 | Rule of Simplification/Sequent Form/Formulation 2 | {{begin-eqn}}
{{eqn | n = 1
| l = \vdash p \land q
| o = \implies
| r = p
}}
{{eqn | n = 2
| l = \vdash p \land q
| o = \implies
| r = q
}}
{{end-eqn}} | 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>$\begin{array}{|ccc|c|c||c|c|} \hline
p & \land & q & p & q & p \land q \implies p & p \land q \implies q \\
\hline
\F & \F & \F & \F & \F & \T & \T \\
\F & \F... | {{begin-eqn}}
{{eqn | n = 1
| l = \vdash p \land q
| o = \implies
| r = p
}}
{{eqn | n = 2
| l = \vdash p \land q
| o = \implies
| r = q
}}
{{end-eqn}} | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{... | Rule of Simplification/Sequent Form/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_by_Truth_Table | [
"Rule of Simplification"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-7037 | Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1 | :$\vdash p \land q \implies p$ | {{BeginTableau|\vdash p \land q \implies p}}
{{Assumption|1|p \land q}}
{{Simplification|2|1|p|1|1}}
{{Implication|3||p \land q \implies p|1|2}}
{{EndTableau|qed}} | :$\vdash p \land q \implies p$ | {{BeginTableau|\vdash p \land q \implies p}}
{{Assumption|1|p \land q}}
{{Simplification|2|1|p|1|1}}
{{Implication|3||p \land q \implies p|1|2}}
{{EndTableau|qed}} | Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 1 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_1/Form_1 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_1/Form_1 | [
"Rule of Simplification"
] | [] | [] |
proofwiki-7038 | Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2 | :$\vdash p \land q \implies q$ | {{BeginTableau|\vdash p \land q \implies q}}
{{Assumption|1|p \land q}}
{{Simplification|2|1|q|1|2}}
{{Implication|3||p \land q \implies q|1|2}}
{{EndTableau}}
{{Qed}} | :$\vdash p \land q \implies q$ | {{BeginTableau|\vdash p \land q \implies q}}
{{Assumption|1|p \land q}}
{{Simplification|2|1|q|1|2}}
{{Implication|3||p \land q \implies q|1|2}}
{{EndTableau}}
{{Qed}} | Rule of Simplification/Sequent Form/Formulation 2/Proof 1/Form 2 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_1/Form_2 | https://proofwiki.org/wiki/Rule_of_Simplification/Sequent_Form/Formulation_2/Proof_1/Form_2 | [
"Rule of Simplification"
] | [] | [] |
proofwiki-7039 | Complement of Complement in Uniquely Complemented Lattice | Let $\struct {S, \wedge, \vee, \preceq}$ be a uniquely complemented lattice.
For each $x \in S$, let $\neg x$ be the complement of $x$.
Then for each $x \in S$:
:$\neg \neg x = x$ | By the definition of a complement of $x$:
:$\neg x \vee x = \top$
:$\neg x \wedge x = \bot$
Since $\vee$ and $\wedge$ are commutative:
:$x \vee \neg x = \top$
:$x \wedge \neg x = \bot$
Thus by the definition of complement, $x$ is a complement of $\neg x$.
By the definition of a uniquely complemented lattice, $x = \neg ... | Let $\struct {S, \wedge, \vee, \preceq}$ be a [[Definition:Uniquely Complemented Lattice|uniquely complemented lattice]].
For each $x \in S$, let $\neg x$ be the [[Definition:Complement (Lattice Theory)|complement]] of $x$.
Then for each $x \in S$:
:$\neg \neg x = x$ | By the definition of a [[Definition:Complement (Lattice Theory)|complement]] of $x$:
:$\neg x \vee x = \top$
:$\neg x \wedge x = \bot$
Since $\vee$ and $\wedge$ are [[Definition:Commutative Operation|commutative]]:
:$x \vee \neg x = \top$
:$x \wedge \neg x = \bot$
Thus by the definition of [[Definition:Complement (... | Complement of Complement in Uniquely Complemented Lattice | https://proofwiki.org/wiki/Complement_of_Complement_in_Uniquely_Complemented_Lattice | https://proofwiki.org/wiki/Complement_of_Complement_in_Uniquely_Complemented_Lattice | [
"Uniquely Complemented Lattices"
] | [
"Definition:Uniquely Complemented Lattice",
"Definition:Complement (Lattice Theory)"
] | [
"Definition:Complement (Lattice Theory)",
"Definition:Commutative/Operation",
"Definition:Complement (Lattice Theory)",
"Definition:Uniquely Complemented Lattice",
"Category:Uniquely Complemented Lattices"
] |
proofwiki-7040 | De Morgan's Laws imply Uniquely Complemented Lattice is Boolean Lattice | Let $\struct {S, \wedge, \vee, \preceq}$ be a uniquely complemented lattice.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\forall p, q \in S: \neg p \vee \neg q {{=}} \neg \paren {p \wedge q}$}}
{{item|(2):|$\forall p, q \in S: \neg p \wedge \neg q {{=}} \neg \paren {p \vee q}$}}
{{item|(3):|$\forall p, q \in S: p \preceq q... | === $(1)$ implies $(2)$ ===
Suppose:
:$\forall p, q \in S: \neg p \vee \neg q = \neg \paren {p \wedge q}$
Then applying this to $\neg p$ and $\neg q$:
:$\neg \neg p \vee \neg \neg q = \neg \paren {\neg p \wedge \neg q}$
By Complement of Complement in Uniquely Complemented Lattice:
:$\neg \neg p = p$ and $\neg \neg q = ... | Let $\struct {S, \wedge, \vee, \preceq}$ be a [[Definition:Uniquely Complemented Lattice|uniquely complemented lattice]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\forall p, q \in S: \neg p \vee \neg q {{=}} \neg \paren {p \wedge q}$}}
{{item|(2):|$\forall p, q \in S: \neg p \wedge \neg q {{=}} \neg \paren {p \vee q}$... | === $(1)$ implies $(2)$ ===
Suppose:
:$\forall p, q \in S: \neg p \vee \neg q = \neg \paren {p \wedge q}$
Then applying this to $\neg p$ and $\neg q$:
:$\neg \neg p \vee \neg \neg q = \neg \paren {\neg p \wedge \neg q}$
By [[Complement of Complement in Uniquely Complemented Lattice]]:
:$\neg \neg p = p$ and $\neg... | De Morgan's Laws imply Uniquely Complemented Lattice is Boolean Lattice | https://proofwiki.org/wiki/De_Morgan's_Laws_imply_Uniquely_Complemented_Lattice_is_Boolean_Lattice | https://proofwiki.org/wiki/De_Morgan's_Laws_imply_Uniquely_Complemented_Lattice_is_Boolean_Lattice | [
"Uniquely Complemented Lattices",
"Boolean Lattices",
"Distributive Lattices"
] | [
"Definition:Uniquely Complemented Lattice",
"Definition:Distributive Lattice",
"Definition:Boolean Lattice"
] | [
"Complement of Complement in Uniquely Complemented Lattice",
"Definition:Complement (Lattice Theory)",
"Complement of Complement in Uniquely Complemented Lattice",
"Definition:Complement (Lattice Theory)",
"Complement of Complement in Uniquely Complemented Lattice",
"Complement of Complement in Uniquely C... |
proofwiki-7041 | Complement of Bottom/Boolean Algebra | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then:
:$\neg \bot = \top$ | Since $\bot$ is the identity for $\vee$, the first condition for $\neg \bot$:
:$\bot \vee \neg \bot = \top$
implies that $\neg \bot = \top$ is the only possibility.
Since $\top$ is the identity for $\wedge$, it follows that:
:$\bot \wedge \top = \bot$
and we conclude that:
:$\neg \bot = \top$
as desired.
{{qed}} | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
Then:
:$\neg \bot = \top$ | Since $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$, the first condition for $\neg \bot$:
:$\bot \vee \neg \bot = \top$
implies that $\neg \bot = \top$ is the only possibility.
Since $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$, it follows that:
:$\bot \wedge \top = \bot$
... | Complement of Bottom/Boolean Algebra | https://proofwiki.org/wiki/Complement_of_Bottom/Boolean_Algebra | https://proofwiki.org/wiki/Complement_of_Bottom/Boolean_Algebra | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-7042 | Complement of Bottom/Bounded Lattice | Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded lattice.
Then the bottom $\bot$ has a unique complement, namely $\top$, top. | We know that $\bot$ is the identity for $\vee$.
Therefore, from the condition that:
:$\bot \vee a = \top$
for a complement $a$ of $\bot$, it follows that $a = \top$ is the only possibility.
Since also:
:$\bot \wedge \top = \bot$
as $\top$ is the identity for $\wedge$, the result follows.
{{qed}} | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Lattice|bounded lattice]].
Then the [[Definition:Bottom of Lattice|bottom]] $\bot$ has a [[Definition:Unique|unique]] [[Definition:Complement (Lattice Theory)|complement]], namely $\top$, [[Definition:Top of Lattice|top]]. | We know that $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$.
Therefore, from the condition that:
:$\bot \vee a = \top$
for a [[Definition:Complement (Lattice Theory)|complement]] $a$ of $\bot$, it follows that $a = \top$ is the only possibility.
Since also:
:$\bot \wedge \top = \bot$
as $\top$ ... | Complement of Bottom/Bounded Lattice | https://proofwiki.org/wiki/Complement_of_Bottom/Bounded_Lattice | https://proofwiki.org/wiki/Complement_of_Bottom/Bounded_Lattice | [
"Bounded Lattices"
] | [
"Definition:Bounded Lattice",
"Definition:Bottom of Lattice",
"Definition:Unique",
"Definition:Complement (Lattice Theory)",
"Definition:Top of Lattice"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Complement (Lattice Theory)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-7043 | Complement of Top/Bounded Lattice | Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded lattice.
Then the top $\top$ has a unique complement, namely $\bot$, bottom. | By Dual Pairs (Order Theory), $\top$ is dual to $\bot$.
The result follows from the Duality Principle and Complement of Bottom.
{{qed}} | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Lattice|bounded lattice]].
Then the [[Definition:Top of Lattice|top]] $\top$ has a [[Definition:Unique|unique]] [[Definition:Complement (Lattice Theory)|complement]], namely $\bot$, [[Definition:Bottom of Lattice|bottom]]. | By [[Dual Pairs (Order Theory)]], $\top$ is [[Definition:Dual Statement (Order Theory)|dual]] to $\bot$.
The result follows from the [[Duality Principle (Order Theory)|Duality Principle]] and [[Complement of Bottom (Bounded Lattice)|Complement of Bottom]].
{{qed}} | Complement of Top/Bounded Lattice | https://proofwiki.org/wiki/Complement_of_Top/Bounded_Lattice | https://proofwiki.org/wiki/Complement_of_Top/Bounded_Lattice | [
"Bounded Lattices"
] | [
"Definition:Bounded Lattice",
"Definition:Top of Lattice",
"Definition:Unique",
"Definition:Complement (Lattice Theory)",
"Definition:Bottom of Lattice"
] | [
"Dual Pairs (Order Theory)",
"Definition:Dual Statement (Order Theory)",
"Duality Principle (Order Theory)",
"Complement of Bottom/Bounded Lattice"
] |
proofwiki-7044 | Complement of Top/Boolean Algebra | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then:
:$\neg \top = \bot$ | Since $\top$ is the identity for $\wedge$, the second condition for $\neg \top$:
:$\top \wedge \neg \top = \bot$
implies that $\neg \top = \bot$ is the only possibility.
Since $\bot$ is the identity for $\vee$, it follows that:
:$\top \vee \bot = \top$
and we conclude that:
:$\neg \top = \bot$
as desired.
{{qed}} | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
Then:
:$\neg \top = \bot$ | Since $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$, the second condition for $\neg \top$:
:$\top \wedge \neg \top = \bot$
implies that $\neg \top = \bot$ is the only possibility.
Since $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$, it follows that:
:$\top \vee \bot = \top$... | Complement of Top/Boolean Algebra | https://proofwiki.org/wiki/Complement_of_Top/Boolean_Algebra | https://proofwiki.org/wiki/Complement_of_Top/Boolean_Algebra | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-7045 | Duality Principle (Boolean Algebras) | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then any theorem in $\struct {S, \vee, \wedge, \neg}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem. | Let us take the axioms of a Boolean algebra $\struct {S, \wedge, \vee, \neg}$:
{{:Axiom:Boolean Algebra/Axioms/Formulation 1}}
It can be seen by inspection that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout, does not change the axioms.
Thus, what you get is a Boolean algebra again.
Hence the result.
... | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
Then any theorem in $\struct {S, \vee, \wedge, \neg}$ remains valid if both $\vee$ and $\wedge$ are interchanged, and also $\bot$ and $\top$ are interchanged throughout the whole theorem. | Let us take the [[Axiom:Boolean Algebra/Axioms/Formulation 1|axioms of a Boolean algebra]] $\struct {S, \wedge, \vee, \neg}$:
{{:Axiom:Boolean Algebra/Axioms/Formulation 1}}
It can be seen by inspection that exchanging $\wedge$ and $\vee$, and $\bot$ and $\top$ throughout, does not change the axioms.
Thus, what you ... | Duality Principle (Boolean Algebras) | https://proofwiki.org/wiki/Duality_Principle_(Boolean_Algebras) | https://proofwiki.org/wiki/Duality_Principle_(Boolean_Algebras) | [
"Boolean Algebras",
"Named Theorems"
] | [
"Definition:Boolean Algebra"
] | [
"Axiom:Boolean Algebra/Axioms/Formulation 1",
"Definition:Boolean Algebra"
] |
proofwiki-7046 | Identities of Boolean Algebra are also Zeroes | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1.
Let the identity for $\vee$ be $\bot$ and the identity for $\wedge$ be $\top$.
Then:
{{begin-eqn}}
{{eqn | n = 1
| q = \forall x \in S
| l = x \vee \top
| r = \top
}}
{{eqn | n = 2
| q = \forall x \in S
... | Let $x \in S$.
Then:
{{begin-eqn}}
{{eqn | l = x \vee \top
| r = \paren {x \vee \top} \wedge \top
| c = {{Boolean-algebra-axiom|1|3}}: $\top$ is the identity of $\wedge$
}}
{{eqn | r = \paren {x \vee \top} \wedge \paren {x \vee \neg x}
| c = {{Boolean-algebra-axiom|1|4}}: $x \vee x' = \top$
}}
{{eqn |... | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra/Definition 1|Boolean algebra, defined as in Definition 1]].
Let the [[Definition:Identity Element|identity]] for $\vee$ be $\bot$ and the [[Definition:Identity Element|identity]] for $\wedge$ be $\top$.
Then:
{{begin-eqn}}
{{eqn | n = 1
|... | Let $x \in S$.
Then:
{{begin-eqn}}
{{eqn | l = x \vee \top
| r = \paren {x \vee \top} \wedge \top
| c = {{Boolean-algebra-axiom|1|3}}: $\top$ is the [[Definition:Identity Element|identity]] of $\wedge$
}}
{{eqn | r = \paren {x \vee \top} \wedge \paren {x \vee \neg x}
| c = {{Boolean-algebra-axiom|1|... | Identities of Boolean Algebra are also Zeroes | https://proofwiki.org/wiki/Identities_of_Boolean_Algebra_are_also_Zeroes | https://proofwiki.org/wiki/Identities_of_Boolean_Algebra_are_also_Zeroes | [
"Identities of Boolean Algebra are also Zeroes",
"Identity Elements",
"Zero Elements",
"Boolean Algebras"
] | [
"Definition:Boolean Algebra/Definition 1",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Zero Element",
"Definition:Zero Element"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Distributive Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Duality Principle (Boolean Algebras)"
] |
proofwiki-7047 | Complement of Complement (Boolean Algebras) | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then for all $a \in S$:
:$\map \neg {\neg a} = a$ | Follows directly from Complement in Boolean Algebra is Unique.
{{qed}} | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
Then for all $a \in S$:
:$\map \neg {\neg a} = a$ | Follows directly from [[Complement in Boolean Algebra is Unique]].
{{qed}} | Complement of Complement (Boolean Algebras) | https://proofwiki.org/wiki/Complement_of_Complement_(Boolean_Algebras) | https://proofwiki.org/wiki/Complement_of_Complement_(Boolean_Algebras) | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra"
] | [
"Complement in Boolean Algebra is Unique"
] |
proofwiki-7048 | De Morgan's Laws (Boolean Algebras) | :$\neg \paren {a \vee b} = \neg a \wedge \neg b$
:$\neg \paren {a \wedge b} = \neg a \vee \neg b$ | By virtue of Complement in Boolean Algebra is Unique, it will suffice to verify:
:$\paren {a \vee b} \wedge \paren {\neg a \wedge \neg b} = \bot$
:$\paren {a \vee b} \vee \paren {\neg a \wedge \neg b} = \top$
For the first of these, compute:
{{begin-eqn}}
{{eqn | l = \paren {a \vee b} \wedge \paren {\neg a \wedge \neg ... | :$\neg \paren {a \vee b} = \neg a \wedge \neg b$
:$\neg \paren {a \wedge b} = \neg a \vee \neg b$ | By virtue of [[Complement in Boolean Algebra is Unique]], it will suffice to verify:
:$\paren {a \vee b} \wedge \paren {\neg a \wedge \neg b} = \bot$
:$\paren {a \vee b} \vee \paren {\neg a \wedge \neg b} = \top$
For the first of these, compute:
{{begin-eqn}}
{{eqn | l = \paren {a \vee b} \wedge \paren {\neg a \wedg... | De Morgan's Laws (Boolean Algebras) | https://proofwiki.org/wiki/De_Morgan's_Laws_(Boolean_Algebras) | https://proofwiki.org/wiki/De_Morgan's_Laws_(Boolean_Algebras) | [
"Boolean Algebras",
"De Morgan's Laws"
] | [] | [
"Complement in Boolean Algebra is Unique",
"Definition:Associative Operation",
"Definition:Distributive Operation",
"Definition:Boolean Algebra",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Identities of Boolean ... |
proofwiki-7049 | Complement in Boolean Algebra is Unique | Let $\left({S, \vee, \wedge}\right)$ be a Boolean algebra.
Then for all $a \in S$, there is a unique $b \in S$ such that:
:$a \wedge b = \bot, a \vee b = \top$
i.e., a valid choice for $\neg a$ as in axiom $(BA \ 4)$ for Boolean algebras. | Suppose $b, c \in S$ both satisfy the identities.
Then:
{{begin-eqn}}
{{eqn | l = b
| r = b \wedge \top
| c = $\top$ is the identity for $\wedge$
}}
{{eqn | r = b \wedge \left({a \vee c}\right)
| c = by hypothesis
}}
{{eqn | r = \left({b \wedge a}\right) \vee \left({b \wedge c}\right)
| c = $\we... | Let $\left({S, \vee, \wedge}\right)$ be a [[Definition:Boolean Algebra|Boolean algebra]].
Then for all $a \in S$, there is a [[Definition:Unique|unique]] $b \in S$ such that:
:$a \wedge b = \bot, a \vee b = \top$
i.e., a valid choice for $\neg a$ as in axiom $(BA \ 4)$ for [[Definition:Boolean Algebra|Boolean algeb... | Suppose $b, c \in S$ both satisfy the identities.
Then:
{{begin-eqn}}
{{eqn | l = b
| r = b \wedge \top
| c = $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
}}
{{eqn | r = b \wedge \left({a \vee c}\right)
| c = by hypothesis
}}
{{eqn | r = \left({b \wedge a}\right) \vee \left({b... | Complement in Boolean Algebra is Unique | https://proofwiki.org/wiki/Complement_in_Boolean_Algebra_is_Unique | https://proofwiki.org/wiki/Complement_in_Boolean_Algebra_is_Unique | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra",
"Definition:Unique",
"Definition:Boolean Algebra"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Distributive Operation",
"Definition:Distributive Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Category:Boolean Algebras"
] |
proofwiki-7050 | Cancellation of Join in Boolean Algebra | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Let $a, b, c \in S$, and suppose that:
{{begin-eqn}}
{{eqn | l = a \vee c
| r = b \vee c
}}
{{eqn | l = a \vee \neg c
| r = b \vee \neg c
}}
{{end-eqn}}
Then $a = b$. | {{begin-eqn}}
{{eqn | l = a
| r = a \vee \bot
| c = {{Boolean-algebra-axiom|1|3}}: $\bot$ is the identity for $\vee$
}}
{{eqn | r = a \vee \paren {c \wedge \neg c}
| c = {{Boolean-algebra-axiom|1|4}}
}}
{{eqn | r = \paren {a \vee c} \wedge \paren {a \vee \neg c}
| c = {{Boolean-algebra-axiom|1|2... | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
Let $a, b, c \in S$, and suppose that:
{{begin-eqn}}
{{eqn | l = a \vee c
| r = b \vee c
}}
{{eqn | l = a \vee \neg c
| r = b \vee \neg c
}}
{{end-eqn}}
Then $a = b$. | {{begin-eqn}}
{{eqn | l = a
| r = a \vee \bot
| c = {{Boolean-algebra-axiom|1|3}}: $\bot$ is the [[Definition:Identity Element|identity]] for $\vee$
}}
{{eqn | r = a \vee \paren {c \wedge \neg c}
| c = {{Boolean-algebra-axiom|1|4}}
}}
{{eqn | r = \paren {a \vee c} \wedge \paren {a \vee \neg c}
|... | Cancellation of Join in Boolean Algebra | https://proofwiki.org/wiki/Cancellation_of_Join_in_Boolean_Algebra | https://proofwiki.org/wiki/Cancellation_of_Join_in_Boolean_Algebra | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Distributive Operation",
"Definition:Distributive Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-7051 | Cancellation of Meet in Boolean Algebra | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Let $a, b, c \in S$.
Let:
{{begin-eqn}}
{{eqn | l = a \wedge c
| r = b \wedge c
}}
{{eqn | l = a \wedge \neg c
| r = b \wedge \neg c
}}
{{end-eqn}}
Then:
: $a = b$ | Follows from Cancellation of Join in Boolean Algebra through the Duality Principle
{{qed}} | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra|Boolean algebra]].
Let $a, b, c \in S$.
Let:
{{begin-eqn}}
{{eqn | l = a \wedge c
| r = b \wedge c
}}
{{eqn | l = a \wedge \neg c
| r = b \wedge \neg c
}}
{{end-eqn}}
Then:
: $a = b$ | Follows from [[Cancellation of Join in Boolean Algebra]] through the [[Duality Principle (Boolean Algebras)|Duality Principle]]
{{qed}} | Cancellation of Meet in Boolean Algebra | https://proofwiki.org/wiki/Cancellation_of_Meet_in_Boolean_Algebra | https://proofwiki.org/wiki/Cancellation_of_Meet_in_Boolean_Algebra | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra"
] | [
"Cancellation of Join in Boolean Algebra",
"Duality Principle (Boolean Algebras)"
] |
proofwiki-7052 | Operations of Boolean Algebra are Associative | Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra, defined as in Definition 1.
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
| l = \paren {a \wedge b} \wedge c
| r = a \wedge \paren {b \wedge c}
}}
{{eqn | q = \forall a, b, c \in S
| l = \paren {a \vee b} \vee c
| r = a \vee \pa... | Let $a, b, c \in S$.
Let:
:$x = a \wedge \paren {b \wedge c}$
:$y = \paren {a \wedge b} \wedge c$
Then:
{{begin-eqn}}
{{eqn | l = a \vee x
| r = a \vee \paren {a \wedge \paren {b \wedge c} }
| c =
}}
{{eqn | r = \paren {a \vee a} \wedge \paren {a \vee \paren {b \wedge c} }
| c = {{Boolean-algebra-axi... | Let $\struct {S, \vee, \wedge, \neg}$ be a [[Definition:Boolean Algebra/Definition 1|Boolean algebra, defined as in Definition 1]].
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b, c \in S
| l = \paren {a \wedge b} \wedge c
| r = a \wedge \paren {b \wedge c}
}}
{{eqn | q = \forall a, b, c \in S
| l = ... | Let $a, b, c \in S$.
Let:
:$x = a \wedge \paren {b \wedge c}$
:$y = \paren {a \wedge b} \wedge c$
Then:
{{begin-eqn}}
{{eqn | l = a \vee x
| r = a \vee \paren {a \wedge \paren {b \wedge c} }
| c =
}}
{{eqn | r = \paren {a \vee a} \wedge \paren {a \vee \paren {b \wedge c} }
| c = {{Boolean-algebra-... | Operations of Boolean Algebra are Associative | https://proofwiki.org/wiki/Operations_of_Boolean_Algebra_are_Associative | https://proofwiki.org/wiki/Operations_of_Boolean_Algebra_are_Associative | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra/Definition 1",
"Definition:Associative Operation"
] | [
"Definition:Distributive Operation",
"Operations of Boolean Algebra are Idempotent",
"Absorption Laws (Boolean Algebras)",
"Definition:Distributive Operation",
"Absorption Laws (Boolean Algebras)",
"Absorption Laws (Boolean Algebras)",
"Definition:Distributive Operation",
"Definition:Identity (Abstrac... |
proofwiki-7053 | Absorption Laws (Boolean Algebras) | Let $\struct {S, \vee, \wedge}$ be a Boolean algebra, defined as in Definition 1.
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b \in S
| l = a
| r = a \vee \paren {a \wedge b}
}}
{{eqn | l = a
| r = a \wedge \paren {a \vee b}
}}
{{end-eqn}}
That is, $\vee$ absorbs $\wedge$, and $\wedge$ absorbs $\vee$. | Let $a, b \in S$.
Then:
{{begin-eqn}}
{{eqn | l = a \vee \paren {a \wedge b}
| r = \paren {a \wedge \top} \vee \paren {a \wedge b}
| c = {{Boolean-algebra-axiom|1|3}}: $\top$ is the identity for $\wedge$
}}
{{eqn | r = a \wedge \paren {\top \vee b}
| c = {{Boolean-algebra-axiom|1|2}}: $\wedge$ distrib... | Let $\struct {S, \vee, \wedge}$ be a [[Definition:Boolean Algebra/Definition 1|Boolean algebra, defined as in Definition 1]].
Then:
{{begin-eqn}}
{{eqn | q = \forall a, b \in S
| l = a
| r = a \vee \paren {a \wedge b}
}}
{{eqn | l = a
| r = a \wedge \paren {a \vee b}
}}
{{end-eqn}}
That is, $\vee$ ... | Let $a, b \in S$.
Then:
{{begin-eqn}}
{{eqn | l = a \vee \paren {a \wedge b}
| r = \paren {a \wedge \top} \vee \paren {a \wedge b}
| c = {{Boolean-algebra-axiom|1|3}}: $\top$ is the [[Definition:Identity Element|identity]] for $\wedge$
}}
{{eqn | r = a \wedge \paren {\top \vee b}
| c = {{Boolean-al... | Absorption Laws (Boolean Algebras) | https://proofwiki.org/wiki/Absorption_Laws_(Boolean_Algebras) | https://proofwiki.org/wiki/Absorption_Laws_(Boolean_Algebras) | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra/Definition 1",
"Definition:Absorption Law",
"Definition:Absorption Law"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Distributive Operation",
"Identities of Boolean Algebra are also Zeroes",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Duality Principle (Boolean Algebras)"
] |
proofwiki-7054 | Difference of Absolutely Convergent Series | Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two real or complex series that are absolutely convergent.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {a_n - b_n}$ is absolutely convergent, and:
:$\ds \sum_{n \mathop = 1}^\infty \paren {a_n - b_n} = \sum_{n \mathop = 1... | The series $\ds \sum_{n \mathop = 1}^\infty \paren {-b_n}$ is absolutely convergent, as $\cmod {-b_n} = \cmod {b_n}$ for all $n \in \N$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n
| r = \sum_{n \mathop = 1}^\infty a_n + \paren {-1} \sum_{n \mathop = 1}^\infty... | Let $\ds \sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ be two [[Definition:Real Number|real]] or [[Definition:Complex Number|complex]] [[Definition:Series|series]] that are [[Definition:Absolutely Convergent Series|absolutely convergent]].
Then the [[Definition:Series|series]] $\ds \sum_{... | The [[Definition:Series|series]] $\ds \sum_{n \mathop = 1}^\infty \paren {-b_n}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]], as $\cmod {-b_n} = \cmod {b_n}$ for all $n \in \N$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty a_n - \sum_{n \mathop = 1}^\infty b_n
| r = \sum_... | Difference of Absolutely Convergent Series | https://proofwiki.org/wiki/Difference_of_Absolutely_Convergent_Series | https://proofwiki.org/wiki/Difference_of_Absolutely_Convergent_Series | [
"Absolute Convergence"
] | [
"Definition:Real Number",
"Definition:Complex Number",
"Definition:Series",
"Definition:Absolutely Convergent Series",
"Definition:Series",
"Definition:Absolutely Convergent Series"
] | [
"Definition:Series",
"Definition:Absolutely Convergent Series",
"Manipulation of Absolutely Convergent Series/Scale Factor",
"Sum of Absolutely Convergent Series"
] |
proofwiki-7055 | True Statement is implied by Every Statement/Formulation 2 | :$\vdash q \implies \paren {p \implies q}$ | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|q}}
{{SequentIntro|2|1|p \implies q|1|True Statement is implied by Every Statement: Formulation 1}}
{{Implication|3||q \implies \paren {p \implies q}|1|2}}
{{EndTableau}}
{{qed}} | :$\vdash q \implies \paren {p \implies q}$ | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|q}}
{{SequentIntro|2|1|p \implies q|1|[[True Statement is implied by Every Statement/Formulation 1|True Statement is implied by Every Statement: Formulation 1]]}}
{{Implication|3||q \implies \paren {p \implies q}|1|2}}
{{EndTableau}}
{{qed}} | True Statement is implied by Every Statement/Formulation 2/Proof 1 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2/Proof_1 | [
"True Statement is implied by Every Statement"
] | [] | [
"True Statement is implied by Every Statement/Formulation 1"
] |
proofwiki-7056 | True Statement is implied by Every Statement/Formulation 2 | :$\vdash q \implies \paren {p \implies q}$ | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|q}}
{{Implication|3|1|p \implies q|1|2}}
{{Implication|4||q \implies \paren {p \implies q}|2|3}}
{{EndTableau}}
{{qed}} | :$\vdash q \implies \paren {p \implies q}$ | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|q}}
{{Implication|3|1|p \implies q|1|2}}
{{Implication|4||q \implies \paren {p \implies q}|2|3}}
{{EndTableau}}
{{qed}} | True Statement is implied by Every Statement/Formulation 2/Proof 2 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2/Proof_2 | [
"True Statement is implied by Every Statement"
] | [] | [] |
proofwiki-7057 | True Statement is implied by Every Statement/Formulation 2 | :$\vdash q \implies \paren {p \implies q}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective, the first instance of $\implies$, is $\T$ for each boolean interpretation.
$\begin{array}{|ccccc|} \hline
q & \implies & ( p & \implies & q ) \\
\hline
\F & \T & \T & \F & \F \\
\F & \T & \F & \T & \F \\
\T & \... | :$\vdash q \implies \paren {p \implies q}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective|main connective]], the first instance of $\implies$, is $\T$ for each [[Definition:Boolean Interpretation|boolean interpretation]].
$\begin{array}{|ccccc|} \hline
q ... | True Statement is implied by Every Statement/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2/Proof_by_Truth_Table | [
"True Statement is implied by Every Statement"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective",
"Definition:Boolean Interpretation"
] |
proofwiki-7058 | True Statement is implied by Every Statement | ==== Formulation 1 ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:True Statement is implied by Every Statement/Formulation 2}} | {{BeginTableau|p \vdash q \implies p}}
{{Premise|1|p}}
{{Addition|2|1|\neg q \lor p|1|2}}
{{SequentIntro|3|1|q \implies p|1|Rule of Material Implication}}
{{EndTableau}}
{{qed}} | ==== [[True Statement is implied by Every Statement/Formulation 1|Formulation 1]] ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== [[True Statement is implied by Every Statement/Formulation 2|Formulation 2]] ====
{{:True Statement is implied by Every Statement/Formulation 2}} | {{BeginTableau|p \vdash q \implies p}}
{{Premise|1|p}}
{{Addition|2|1|\neg q \lor p|1|2}}
{{SequentIntro|3|1|q \implies p|1|[[Rule of Material Implication/Formulation 1/Reverse Implication|Rule of Material Implication]]}}
{{EndTableau}}
{{qed}} | True Statement is implied by Every Statement/Formulation 1/Proof 1 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_1/Proof_1 | [
"True Statement is implied by Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"True Statement is implied by Every Statement/Formulation 1",
"True Statement is implied by Every Statement/Formulation 2"
] | [
"Rule of Material Implication/Formulation 1/Reverse Implication"
] |
proofwiki-7059 | True Statement is implied by Every Statement | ==== Formulation 1 ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:True Statement is implied by Every Statement/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, where the truth value in the relevant column on the {{LHS}} is $\T$, that under the one on the {{RHS}} is also $\T$:
:<nowiki>$\begin{array}{|c||ccc|} \hline
p & q & \implies & p \\
\hline
\F & \F & \T & \F \\
\F & \T & \F & \F \\
\T & \F & \T & \T \\
\... | ==== [[True Statement is implied by Every Statement/Formulation 1|Formulation 1]] ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== [[True Statement is implied by Every Statement/Formulation 2|Formulation 2]] ====
{{:True Statement is implied by Every Statement/Formulation 2}} | We apply the [[Method of Truth Tables]].
As can be seen by inspection, where the [[Definition:Truth Value|truth value]] in the relevant column on the {{LHS}} is $\T$, that under the one on the {{RHS}} is also $\T$:
:<nowiki>$\begin{array}{|c||ccc|} \hline
p & q & \implies & p \\
\hline
\F & \F & \T & \F \\
\F & \T & ... | True Statement is implied by Every Statement/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_1/Proof_by_Truth_Table | [
"True Statement is implied by Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"True Statement is implied by Every Statement/Formulation 1",
"True Statement is implied by Every Statement/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value"
] |
proofwiki-7060 | True Statement is implied by Every Statement | ==== Formulation 1 ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:True Statement is implied by Every Statement/Formulation 2}} | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|q}}
{{SequentIntro|2|1|p \implies q|1|True Statement is implied by Every Statement: Formulation 1}}
{{Implication|3||q \implies \paren {p \implies q}|1|2}}
{{EndTableau}}
{{qed}} | ==== [[True Statement is implied by Every Statement/Formulation 1|Formulation 1]] ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== [[True Statement is implied by Every Statement/Formulation 2|Formulation 2]] ====
{{:True Statement is implied by Every Statement/Formulation 2}} | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|q}}
{{SequentIntro|2|1|p \implies q|1|[[True Statement is implied by Every Statement/Formulation 1|True Statement is implied by Every Statement: Formulation 1]]}}
{{Implication|3||q \implies \paren {p \implies q}|1|2}}
{{EndTableau}}
{{qed}} | True Statement is implied by Every Statement/Formulation 2/Proof 1 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2/Proof_1 | [
"True Statement is implied by Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"True Statement is implied by Every Statement/Formulation 1",
"True Statement is implied by Every Statement/Formulation 2"
] | [
"True Statement is implied by Every Statement/Formulation 1"
] |
proofwiki-7061 | True Statement is implied by Every Statement | ==== Formulation 1 ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:True Statement is implied by Every Statement/Formulation 2}} | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|q}}
{{Implication|3|1|p \implies q|1|2}}
{{Implication|4||q \implies \paren {p \implies q}|2|3}}
{{EndTableau}}
{{qed}} | ==== [[True Statement is implied by Every Statement/Formulation 1|Formulation 1]] ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== [[True Statement is implied by Every Statement/Formulation 2|Formulation 2]] ====
{{:True Statement is implied by Every Statement/Formulation 2}} | {{BeginTableau|\vdash q \implies \paren {p \implies q} }}
{{Assumption|1|p}}
{{Assumption|2|q}}
{{Implication|3|1|p \implies q|1|2}}
{{Implication|4||q \implies \paren {p \implies q}|2|3}}
{{EndTableau}}
{{qed}} | True Statement is implied by Every Statement/Formulation 2/Proof 2 | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2/Proof_2 | [
"True Statement is implied by Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"True Statement is implied by Every Statement/Formulation 1",
"True Statement is implied by Every Statement/Formulation 2"
] | [] |
proofwiki-7062 | True Statement is implied by Every Statement | ==== Formulation 1 ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:True Statement is implied by Every Statement/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, the truth value under the main connective, the first instance of $\implies$, is $\T$ for each boolean interpretation.
$\begin{array}{|ccccc|} \hline
q & \implies & ( p & \implies & q ) \\
\hline
\F & \T & \T & \F & \F \\
\F & \T & \F & \T & \F \\
\T & \... | ==== [[True Statement is implied by Every Statement/Formulation 1|Formulation 1]] ====
{{:True Statement is implied by Every Statement/Formulation 1}}
==== [[True Statement is implied by Every Statement/Formulation 2|Formulation 2]] ====
{{:True Statement is implied by Every Statement/Formulation 2}} | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective|main connective]], the first instance of $\implies$, is $\T$ for each [[Definition:Boolean Interpretation|boolean interpretation]].
$\begin{array}{|ccccc|} \hline
q ... | True Statement is implied by Every Statement/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement | https://proofwiki.org/wiki/True_Statement_is_implied_by_Every_Statement/Formulation_2/Proof_by_Truth_Table | [
"True Statement is implied by Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"True Statement is implied by Every Statement/Formulation 1",
"True Statement is implied by Every Statement/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective",
"Definition:Boolean Interpretation"
] |
proofwiki-7063 | False Statement implies Every Statement | ==== Formulation 1 ====
{{:False Statement implies Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:False Statement implies Every Statement/Formulation 2}} | {{BeginTableau|\neg p \vdash p \implies q}}
{{Premise|1|\neg p}}
{{Addition|2|1|\neg p \lor q|1|1}}
{{SequentIntro|3|1|p \implies q|2|Rule of Material Implication}}
{{EndTableau|qed}} | ==== [[False Statement implies Every Statement/Formulation 1|Formulation 1]] ====
{{:False Statement implies Every Statement/Formulation 1}}
==== [[False Statement implies Every Statement/Formulation 2|Formulation 2]] ====
{{:False Statement implies Every Statement/Formulation 2}} | {{BeginTableau|\neg p \vdash p \implies q}}
{{Premise|1|\neg p}}
{{Addition|2|1|\neg p \lor q|1|1}}
{{SequentIntro|3|1|p \implies q|2|[[Rule of Material Implication/Formulation 1/Reverse Implication|Rule of Material Implication]]}}
{{EndTableau|qed}} | False Statement implies Every Statement/Formulation 1/Proof 1 | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_1/Proof_1 | [
"False Statement implies Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"False Statement implies Every Statement/Formulation 1",
"False Statement implies Every Statement/Formulation 2"
] | [
"Rule of Material Implication/Formulation 1/Reverse Implication"
] |
proofwiki-7064 | False Statement implies Every Statement | ==== Formulation 1 ====
{{:False Statement implies Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:False Statement implies Every Statement/Formulation 2}} | We apply the Method of Truth Tables.
As can be seen by inspection, where the truth value in the relevant column on the {{LHS}} is $\T$, that under the one on the {{RHS}} is also $\T$:
:<nowiki>$\begin{array}{|cc||ccc|} \hline
\neg & p & p & \implies & q \\
\hline
\T & \F & \F & \T & \F \\
\T & \F & \F & \T & \T \\
\F &... | ==== [[False Statement implies Every Statement/Formulation 1|Formulation 1]] ====
{{:False Statement implies Every Statement/Formulation 1}}
==== [[False Statement implies Every Statement/Formulation 2|Formulation 2]] ====
{{:False Statement implies Every Statement/Formulation 2}} | We apply the [[Method of Truth Tables]].
As can be seen by inspection, where the [[Definition:Truth Value|truth value]] in the relevant column on the {{LHS}} is $\T$, that under the one on the {{RHS}} is also $\T$:
:<nowiki>$\begin{array}{|cc||ccc|} \hline
\neg & p & p & \implies & q \\
\hline
\T & \F & \F & \T & \F ... | False Statement implies Every Statement/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_1/Proof_by_Truth_Table | [
"False Statement implies Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"False Statement implies Every Statement/Formulation 1",
"False Statement implies Every Statement/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value"
] |
proofwiki-7065 | False Statement implies Every Statement | ==== Formulation 1 ====
{{:False Statement implies Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:False Statement implies Every Statement/Formulation 2}} | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{Assumption|2|p}}
{{NonContradiction|3|1, 2|2|1}}
{{Explosion|4|1, 2|q|3}}
{{Implication|5|1|p \implies q|2|4}}
{{Implication|6||\neg p \implies \left({p \implies q}\right)|1|5}}
{{EndTableau}}
{{qed}} | ==== [[False Statement implies Every Statement/Formulation 1|Formulation 1]] ====
{{:False Statement implies Every Statement/Formulation 1}}
==== [[False Statement implies Every Statement/Formulation 2|Formulation 2]] ====
{{:False Statement implies Every Statement/Formulation 2}} | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{Assumption|2|p}}
{{NonContradiction|3|1, 2|2|1}}
{{Explosion|4|1, 2|q|3}}
{{Implication|5|1|p \implies q|2|4}}
{{Implication|6||\neg p \implies \left({p \implies q}\right)|1|5}}
{{EndTableau}}
{{qed}} | False Statement implies Every Statement/Formulation 2/Proof 1 | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_2/Proof_1 | [
"False Statement implies Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"False Statement implies Every Statement/Formulation 1",
"False Statement implies Every Statement/Formulation 2"
] | [] |
proofwiki-7066 | False Statement implies Every Statement | ==== Formulation 1 ====
{{:False Statement implies Every Statement/Formulation 1}}
==== Formulation 2 ====
{{:False Statement implies Every Statement/Formulation 2}} | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{SequentIntro|2|1|p \implies q|1|False Statement implies Every Statement: Formulation 1}}
{{Implication|3||\neg p \implies \left({p \implies q}\right)|1|2}}
{{EndTableau}}
{{qed}} | ==== [[False Statement implies Every Statement/Formulation 1|Formulation 1]] ====
{{:False Statement implies Every Statement/Formulation 1}}
==== [[False Statement implies Every Statement/Formulation 2|Formulation 2]] ====
{{:False Statement implies Every Statement/Formulation 2}} | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{SequentIntro|2|1|p \implies q|1|[[False Statement implies Every Statement/Formulation 1|False Statement implies Every Statement: Formulation 1]]}}
{{Implication|3||\neg p \implies \left({p \implies q}\right)|1|2}}
{{EndTableau... | False Statement implies Every Statement/Formulation 2/Proof 2 | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_2/Proof_2 | [
"False Statement implies Every Statement",
"Paradoxes of Material Implication",
"Conditional"
] | [
"False Statement implies Every Statement/Formulation 1",
"False Statement implies Every Statement/Formulation 2"
] | [
"False Statement implies Every Statement/Formulation 1"
] |
proofwiki-7067 | False Statement implies Every Statement/Formulation 2 | :$\vdash \neg p \implies \paren {p \implies q}$ | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{Assumption|2|p}}
{{NonContradiction|3|1, 2|2|1}}
{{Explosion|4|1, 2|q|3}}
{{Implication|5|1|p \implies q|2|4}}
{{Implication|6||\neg p \implies \left({p \implies q}\right)|1|5}}
{{EndTableau}}
{{qed}} | :$\vdash \neg p \implies \paren {p \implies q}$ | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{Assumption|2|p}}
{{NonContradiction|3|1, 2|2|1}}
{{Explosion|4|1, 2|q|3}}
{{Implication|5|1|p \implies q|2|4}}
{{Implication|6||\neg p \implies \left({p \implies q}\right)|1|5}}
{{EndTableau}}
{{qed}} | False Statement implies Every Statement/Formulation 2/Proof 1 | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_2 | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_2/Proof_1 | [
"False Statement implies Every Statement"
] | [] | [] |
proofwiki-7068 | False Statement implies Every Statement/Formulation 2 | :$\vdash \neg p \implies \paren {p \implies q}$ | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{SequentIntro|2|1|p \implies q|1|False Statement implies Every Statement: Formulation 1}}
{{Implication|3||\neg p \implies \left({p \implies q}\right)|1|2}}
{{EndTableau}}
{{qed}} | :$\vdash \neg p \implies \paren {p \implies q}$ | {{BeginTableau|\vdash \neg p \implies \left({p \implies q}\right)}}
{{Assumption|1|\neg p}}
{{SequentIntro|2|1|p \implies q|1|[[False Statement implies Every Statement/Formulation 1|False Statement implies Every Statement: Formulation 1]]}}
{{Implication|3||\neg p \implies \left({p \implies q}\right)|1|2}}
{{EndTableau... | False Statement implies Every Statement/Formulation 2/Proof 2 | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_2 | https://proofwiki.org/wiki/False_Statement_implies_Every_Statement/Formulation_2/Proof_2 | [
"False Statement implies Every Statement"
] | [] | [
"False Statement implies Every Statement/Formulation 1"
] |
proofwiki-7069 | Smullyan's Drinking Principle | Suppose that there is at least one person in the pub.
Then there is a person $x$ in the pub such that if $x$ is drinking, then everyone in the pub is drinking. | We have two choices:
:$\forall y : \map D y$
and
:$\neg \forall y : \map D y$
Suppose $\forall y : \map D y$.
By True Statement is implied by Every Statement:
:$\map D x \implies \forall y : \map D y$
By Existential Generalisation:
:$\exists x : \paren {\map D x \implies \forall y : \map D y}$
Now suppose:
:$\neg \fora... | Suppose that there is at least one person in the pub.
Then there is a person $x$ in the pub such that if $x$ is drinking, then everyone in the pub is drinking. | We have two choices:
:$\forall y : \map D y$
and
:$\neg \forall y : \map D y$
Suppose $\forall y : \map D y$.
By [[True Statement is implied by Every Statement]]:
:$\map D x \implies \forall y : \map D y$
By [[Existential Generalisation]]:
:$\exists x : \paren {\map D x \implies \forall y : \map D y}$
Now suppose... | Smullyan's Drinking Principle/Formal Proof | https://proofwiki.org/wiki/Smullyan's_Drinking_Principle | https://proofwiki.org/wiki/Smullyan's_Drinking_Principle/Formal_Proof | [
"Veridical Paradoxes",
"Logic",
"Smullyan's Drinking Principle"
] | [] | [
"True Statement is implied by Every Statement",
"Existential Generalisation",
"De Morgan's Laws (Predicate Logic)/Denial of Universality",
"False Statement implies Every Statement",
"Existential Generalisation"
] |
proofwiki-7070 | Smullyan's Drinking Principle | Suppose that there is at least one person in the pub.
Then there is a person $x$ in the pub such that if $x$ is drinking, then everyone in the pub is drinking. | Either everyone in the pub is drinking or someone in the pub is not drinking.
Suppose that everyone in the pub is drinking.
By True Statement is implied by Every Statement, the statement:
:''everyone in the pub is drinking''
is implied by the statement:
:''$x$ is drinking''
for any $x$ in the pub.
Since the pub is by a... | Suppose that there is at least one person in the pub.
Then there is a person $x$ in the pub such that if $x$ is drinking, then everyone in the pub is drinking. | Either everyone in the pub is drinking or someone in the pub is not drinking.
Suppose that everyone in the pub is drinking.
By [[True Statement is implied by Every Statement]], the [[Definition:Statement|statement]]:
:''everyone in the pub is drinking''
is implied by the [[Definition:Statement|statement]]:
:''$x$ is... | Smullyan's Drinking Principle/Semi-Formal Proof | https://proofwiki.org/wiki/Smullyan's_Drinking_Principle | https://proofwiki.org/wiki/Smullyan's_Drinking_Principle/Semi-Formal_Proof | [
"Veridical Paradoxes",
"Logic",
"Smullyan's Drinking Principle"
] | [] | [
"True Statement is implied by Every Statement",
"Definition:Statement",
"Definition:Statement",
"Definition:Non-Empty Set",
"False Statement implies Every Statement"
] |
proofwiki-7071 | Two is Boolean Algebra | Let $\mathbf 2$ denote two.
Then $\mathbf 2$ is a Boolean algebra. | It is useful to first state the Cayley tables for the three logical operations $\lor$, $\land$ and $\neg$:
:<nowiki>$\begin{array}{c|cc}
\lor & \bot & \top \\ \hline
\bot & \bot & \top \\
\top & \top & \top
\end{array} \qquad \begin{array}{c|cc}
\land & \bot & \top \\ \hline
\bot & \bot & \bot \\
\top & \bot & ... | Let $\mathbf 2$ denote [[Definition:Two (Boolean Algebra)|two]].
Then $\mathbf 2$ is a [[Definition:Boolean Algebra|Boolean algebra]]. | It is useful to first state the [[Definition:Cayley Table|Cayley tables]] for the three logical operations $\lor$, $\land$ and $\neg$:
:<nowiki>$\begin{array}{c|cc}
\lor & \bot & \top \\ \hline
\bot & \bot & \top \\
\top & \top & \top
\end{array} \qquad \begin{array}{c|cc}
\land & \bot & \top \\ \hline
\bot & \b... | Two is Boolean Algebra | https://proofwiki.org/wiki/Two_is_Boolean_Algebra | https://proofwiki.org/wiki/Two_is_Boolean_Algebra | [
"Boolean Algebras"
] | [
"Definition:Two (Boolean Algebra)",
"Definition:Boolean Algebra"
] | [
"Definition:Cayley Table",
"Definition:Boolean Algebra",
"Definition:Cayley Table",
"Definition:Boolean Algebra"
] |
proofwiki-7072 | Two-Valued Functions form Boolean Algebra | Let $\mathbf 2$ be the Boolean algebra two, and let $X$ be a set.
Let $\mathbf 2^X$ be the set of all mappings $p: X \to \mathbf 2$.
Define the operations $\vee$, $\wedge$ and $\neg$ on $\mathbf 2^X$ in pointwise fashion thus:
:$\vee: \mathbf 2^X \times \mathbf 2^X \to \mathbf 2^X, \left({p \vee q}\right) (x) := p (x) ... | Let us verify the axioms for a Boolean algebra in turn. | Let $\mathbf 2$ be the [[Definition:Boolean Algebra|Boolean algebra]] [[Definition:Two (Boolean Algebra)|two]], and let $X$ be a [[Definition:Set|set]].
Let $\mathbf 2^X$ be the [[Definition:Set of All Mappings|set of all mappings]] $p: X \to \mathbf 2$.
Define the operations $\vee$, $\wedge$ and $\neg$ on $\mathbf ... | Let us verify the axioms for a [[Definition:Boolean Algebra|Boolean algebra]] in turn. | Two-Valued Functions form Boolean Algebra | https://proofwiki.org/wiki/Two-Valued_Functions_form_Boolean_Algebra | https://proofwiki.org/wiki/Two-Valued_Functions_form_Boolean_Algebra | [
"Boolean Algebras"
] | [
"Definition:Boolean Algebra",
"Definition:Two (Boolean Algebra)",
"Definition:Set",
"Definition:Set of All Mappings",
"Definition:Pointwise Operation",
"Definition:Constant Mapping",
"Definition:Boolean Algebra",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Boolean Algebra",
"Definition:Boolean Algebra"
] |
proofwiki-7073 | Power Set and Two-Valued Functions are Isomorphic Boolean Rings | Let $S$ be a set.
Let $\mathbf 2$ be the Boolean ring two.
Let $\powerset S$ be the power set of $S$; by Symmetric Difference with Intersection forms Boolean Ring, it is a Boolean ring.
Let $\mathbf 2^S$ be the set of all mappings $f: S \to \mathbf 2$; by Two-Valued Functions form Boolean Ring, it is also a Boolean rin... | From Support Operation Inverse to Characteristic Function Operation, $\chi_{\paren \cdot}$ is a bijection.
It therefore suffices to establish it is a ring homomorphism.
By Characteristic Function of Symmetric Difference:
:$\chi_{A * B} = \chi_A + \chi_B - 2 \chi_A \chi_B$
Since $\mathbf 2^S$ is a Boolean ring, by Idemp... | Let $S$ be a [[Definition:Set|set]].
Let $\mathbf 2$ be the [[Definition:Boolean Ring|Boolean ring]] [[Definition:Two Ring|two]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$; by [[Symmetric Difference with Intersection forms Boolean Ring]], it is a [[Definition:Boolean Ring|Boolean ring]].
Let... | From [[Support Operation Inverse to Characteristic Function Operation]], $\chi_{\paren \cdot}$ is a [[Definition:Bijection|bijection]].
It therefore suffices to establish it is a [[Definition:Ring Homomorphism|ring homomorphism]].
By [[Characteristic Function of Symmetric Difference]]:
:$\chi_{A * B} = \chi_A + \ch... | Power Set and Two-Valued Functions are Isomorphic Boolean Rings | https://proofwiki.org/wiki/Power_Set_and_Two-Valued_Functions_are_Isomorphic_Boolean_Rings | https://proofwiki.org/wiki/Power_Set_and_Two-Valued_Functions_are_Isomorphic_Boolean_Rings | [
"Power Set",
"Boolean Rings"
] | [
"Definition:Set",
"Definition:Boolean Ring",
"Definition:Two Ring",
"Definition:Power Set",
"Symmetric Difference with Intersection forms Boolean Ring",
"Definition:Boolean Ring",
"Definition:Set of All Mappings",
"Two-Valued Functions form Boolean Ring",
"Definition:Boolean Ring",
"Definition:Cha... | [
"Support Operation Inverse to Characteristic Function Operation",
"Definition:Bijection",
"Definition:Ring Homomorphism",
"Characteristic Function of Symmetric Difference",
"Definition:Boolean Ring",
"Idempotent Ring has Characteristic Two",
"Definition:Ring (Abstract Algebra)/Addition",
"Characterist... |
proofwiki-7074 | Peirce's Law/Formulation 1 | :$\paren {p \implies q} \implies p \vdash p$ | {{BeginTableau|\paren {p \implies q} \implies p \vdash p}}
{{Premise|1|\paren {p \implies q} \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p}}
{{SequentIntro|4|3|p \implies q|3|False Statement implies Every Statement}}
{{ModusPonens|5|1, 3|p|1|4}}
{{Assumption|6|p}}
{{ProofByCases|7|1|p|2|3|5|6|6}... | :$\paren {p \implies q} \implies p \vdash p$ | {{BeginTableau|\paren {p \implies q} \implies p \vdash p}}
{{Premise|1|\paren {p \implies q} \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p}}
{{SequentIntro|4|3|p \implies q|3|[[False Statement implies Every Statement]]}}
{{ModusPonens|5|1, 3|p|1|4}}
{{Assumption|6|p}}
{{ProofByCases|7|1|p|2|3|5|... | Peirce's Law/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_1 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_1/Proof_1 | [
"Peirce's Law"
] | [] | [
"False Statement implies Every Statement"
] |
proofwiki-7075 | Peirce's Law/Formulation 1 | :$\paren {p \implies q} \implies p \vdash p$ | {{BeginTableau|\paren {p \implies q} \implies p \vdash p}}
{{Premise|1|\paren {p \implies q} \implies p}}
{{Assumption|2|\neg p}}
{{SequentIntro|3|2|p \implies q|2|False Statement implies Every Statement}}
{{ModusPonens|4|1,2|p|1|3}}
{{NonContradiction|5|1,2|2|4}}
{{Reductio|6|1|p|2|5}}
{{EndTableau|qed}}
{{LEM|Reducti... | :$\paren {p \implies q} \implies p \vdash p$ | {{BeginTableau|\paren {p \implies q} \implies p \vdash p}}
{{Premise|1|\paren {p \implies q} \implies p}}
{{Assumption|2|\neg p}}
{{SequentIntro|3|2|p \implies q|2|[[False Statement implies Every Statement]]}}
{{ModusPonens|4|1,2|p|1|3}}
{{NonContradiction|5|1,2|2|4}}
{{Reductio|6|1|p|2|5}}
{{EndTableau|qed}}
{{LEM|Re... | Peirce's Law/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_1 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_1/Proof_2 | [
"Peirce's Law"
] | [] | [
"False Statement implies Every Statement"
] |
proofwiki-7076 | Peirce's Law/Formulation 2 | :$\vdash \paren {\paren {p \implies q} \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \implies p} \implies p}}
{{Assumption|1|\paren {p \implies q} \implies p}}
{{SequentIntro|2|1|p|1|Peirce's Law: Formulation 1: $\paren {p \implies q} \implies p \vdash p$}}
{{Implication|3||\paren {\paren {p \implies q} \implies p} \implies p|1|2}}
{{EndTableau}}
{{qe... | :$\vdash \paren {\paren {p \implies q} \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \implies p} \implies p}}
{{Assumption|1|\paren {p \implies q} \implies p}}
{{SequentIntro|2|1|p|1|[[Peirce's Law/Formulation 1|Peirce's Law: Formulation 1]]: $\paren {p \implies q} \implies p \vdash p$}}
{{Implication|3||\paren {\paren {p \implies q} \implies p} \impl... | Peirce's Law/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_2 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_2/Proof_1 | [
"Peirce's Law"
] | [] | [
"Peirce's Law/Formulation 1"
] |
proofwiki-7077 | Peirce's Law/Formulation 2 | :$\vdash \paren {\paren {p \implies q} \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \implies p} \implies p}}
{{Premise|1|\paren {p \implies q} \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p}}
{{SequentIntro|4|3|p \implies q|3|False Statement implies Every Statement}}
{{ModusPonens|5|1, 3|p|1|4}}
{{Assumption|6|p}}
{{ProofByCase... | :$\vdash \paren {\paren {p \implies q} \implies p} \implies p$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \implies p} \implies p}}
{{Premise|1|\paren {p \implies q} \implies p}}
{{ExcludedMiddle|2|p \lor \neg p}}
{{Assumption|3|\neg p}}
{{SequentIntro|4|3|p \implies q|3|[[False Statement implies Every Statement]]}}
{{ModusPonens|5|1, 3|p|1|4}}
{{Assumption|6|p}}
{{ProofBy... | Peirce's Law/Formulation 2/Proof 2 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_2 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_2/Proof_2 | [
"Peirce's Law"
] | [] | [
"False Statement implies Every Statement"
] |
proofwiki-7078 | Peirce's Law/Formulation 2 | :$\vdash \paren {\paren {p \implies q} \implies p} \implies p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective are $\T$ for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc|c|c|}\hline
((p & \implies & q) & \implies & p) & \implies & p \\
\hline
\F & \T & \F & \F & \F & \T & \F \\
\F & \T &... | :$\vdash \paren {\paren {p \implies q} \implies p} \implies 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 connective]] are $\T$ for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|... | Peirce's Law/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Peirce's_Law/Formulation_2 | https://proofwiki.org/wiki/Peirce's_Law/Formulation_2/Proof_by_Truth_Table | [
"Peirce's Law"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7079 | Peirce's Law/Strong Form/Formulation 2 | :$\vdash \paren {\paren {p \implies q} \implies p} \iff p$ | === $(1):$ $\vdash$ Direction ===
{{:Peirce's Law/Strong Form/Formulation 2/Forward Direction}} | :$\vdash \paren {\paren {p \implies q} \implies p} \iff p$ | === $(1):$ [[Peirce's Law/Strong Form/Formulation 2/Forward Direction|$\vdash$ Direction]] ===
{{:Peirce's Law/Strong Form/Formulation 2/Forward Direction}} | Peirce's Law/Strong Form/Formulation 2 | https://proofwiki.org/wiki/Peirce's_Law/Strong_Form/Formulation_2 | https://proofwiki.org/wiki/Peirce's_Law/Strong_Form/Formulation_2 | [
"Peirce's Law"
] | [] | [
"Peirce's Law/Strong Form/Formulation 2/Forward Direction"
] |
proofwiki-7080 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1 | :$p \land \neg q \dashv \vdash \neg \paren {p \implies q}$ | 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>$\begin{array}{|cccc||cccc|} \hline
p & \land & \neg & q & \neg & (p & \implies & q) \\
\hline
\F & \F & \T & \F & \F & \F & \T & \F \\
\F & \F & \F & \T & \F & ... | :$p \land \neg q \dashv \vdash \neg \paren {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>$\begin{array}{|cccc||cccc|} \hline
p... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_1 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_1/Proof_by_Truth_Table | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7081 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Forward Implication | :$p \land \neg q \vdash \neg \paren {p \implies q}$ | {{BeginTableau|p \land \neg q \vdash \neg \paren {p \implies q} }}
{{Premise|1|p \land \neg q}}
{{Assumption|2|p \implies q| Assume the opposite of what is to be proved ...}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|\neg q|1|2}}
{{ModusPonens|5|1, 2|q|2|3}}
{{NonContradiction|6|1, 2|5|4|... and demonstrate a c... | :$p \land \neg q \vdash \neg \paren {p \implies q}$ | {{BeginTableau|p \land \neg q \vdash \neg \paren {p \implies q} }}
{{Premise|1|p \land \neg q}}
{{Assumption|2|p \implies q| Assume the opposite of what is to be proved ...}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|\neg q|1|2}}
{{ModusPonens|5|1, 2|q|2|3}}
{{NonContradiction|6|1, 2|5|4|... and demonstrate a c... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_1/Forward_Implication | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Category:Conjunction with Negative is Equivalent to Negation of Conditional"
] |
proofwiki-7082 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Reverse Implication | :$\neg \paren {p \implies q} \vdash p \land \neg q$ | {{BeginTableau|\neg \paren {p \implies q} \vdash p \land \neg q}}
{{Premise|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|Conditional is Equivalent to Negation of Conjunction with Negative}}
{{NonContradiction|4|1, 2|3|1}}
{{Reductio|5|1|p \land \neg q|... | :$\neg \paren {p \implies q} \vdash p \land \neg q$ | {{BeginTableau|\neg \paren {p \implies q} \vdash p \land \neg q}}
{{Premise|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|[[Conditional is Equivalent to Negation of Conjunction with Negative]]}}
{{NonContradiction|4|1, 2|3|1}}
{{Reductio|5|1|p \land \ne... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_1/Reverse_Implication | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Conditional is Equivalent to Negation of Conjunction with Negative",
"Category:Conjunction with Negative is Equivalent to Negation of Conditional"
] |
proofwiki-7083 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2 | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} } }}
{{Assumption|1|p \land \neg q}}
{{Assumption|2|p \implies q| Assume the opposite of what is to be proved ...}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|\neg q|1|2}}
{{ModusPonens|5|1, 2|q|2|3}}
{{NonContradiction|6|1... | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {p \land \neg q} \implies \paren {\neg \paren {p \implies q} } }}
{{Assumption|1|p \land \neg q}}
{{Assumption|2|p \implies q| Assume the opposite of what is to be proved ...}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|\neg q|1|2}}
{{ModusPonens|5|1, 2|q|2|3}}
{{NonContradiction|6|1... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Forward Implication/Proof 1 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Forward_Implication/Proof_1 | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [] |
proofwiki-7084 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2 | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} } }}
{{Assumption|1|p \land \neg q}}
{{SequentIntro|2|1|\neg \paren {p \implies q}|1|Conjunction with Negative is Equivalent to Negation of Conditional: Formulation 1: Forward Implication}}
{{Implication|3||\paren {p \land \neg q} \im... | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} } }}
{{Assumption|1|p \land \neg q}}
{{SequentIntro|2|1|\neg \paren {p \implies q}|1|[[Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Forward Implication|Conjunction with Negative is Equivalent to Neg... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Proof_1 | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Forward Implication",
"Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Reverse Implication"
] |
proofwiki-7085 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2 | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | 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>$\begin{array}{|cccc|c|cccc|} \hline
p & \land & \neg & q & \iff & \neg & (p & \implies & q) \\
\hline
\F & \F & \T & \F & \T & \F & \F & \T & \F \\
\F & \F & \... | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {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]] is [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Proof_by_Truth_Table | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-7086 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2 | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|Conditional is Equivalent to Negation of Conjunction with Negative}}
{{NonContradiction|4|1, 2|3|1}... | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{Assumption|2|\neg \paren {p \land \neg q} }}
{{SequentIntro|3|2|p \implies q|2|[[Conditional is Equivalent to Negation of Conjunction with Negative]]}}
{{NonContradiction|4|1, 2|... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Reverse_Implication/Proof_1 | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Conditional is Equivalent to Negation of Conjunction with Negative"
] |
proofwiki-7087 | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2 | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|Conjunction with Negative is Equivalent to Negation of Conditional: Formulation 1: Reverse Implication}}
{{Implication|3||\paren {\neg \paren {p... | :$\vdash \paren {p \land \neg q} \iff \paren {\neg \paren {p \implies q} }$ | {{BeginTableau|\vdash \paren {\neg \paren {p \implies q} } \implies \paren {p \land \neg q} }}
{{Assumption|1|\neg \paren {p \implies q} }}
{{SequentIntro|2|1|p \land \neg q|1|[[Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Reverse Implication|Conjunction with Negative is Equivalent t... | Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 2/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2 | https://proofwiki.org/wiki/Conjunction_with_Negative_is_Equivalent_to_Negation_of_Conditional/Formulation_2/Reverse_Implication/Proof_2 | [
"Conjunction with Negative is Equivalent to Negation of Conditional"
] | [] | [
"Conjunction with Negative is Equivalent to Negation of Conditional/Formulation 1/Reverse Implication"
] |
proofwiki-7088 | Reflexive Circular Relation is Equivalence | Let $\RR \subseteq S \times S$ be a reflexive and circular relation in $S$.
Then $\RR$ is an equivalence relation. | To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.
So, checking in turn each of the criteria for equivalence: | Let $\RR \subseteq S \times S$ be a [[Definition:Reflexive Relation|reflexive]] and [[Definition:Circular Relation|circular relation]] in $S$.
Then $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]]. | To prove a [[Definition:Relation|relation]] is an [[Definition:Equivalence Relation|equivalence]], we need to prove it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
So, checking in turn each of the criteria for [[Definitio... | Reflexive Circular Relation is Equivalence | https://proofwiki.org/wiki/Reflexive_Circular_Relation_is_Equivalence | https://proofwiki.org/wiki/Reflexive_Circular_Relation_is_Equivalence | [
"Equivalence Relations",
"Circular Relations",
"Reflexive Relations"
] | [
"Definition:Reflexive Relation",
"Definition:Circular Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Defin... |
proofwiki-7089 | Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result | Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ and $\struct {T, *_1, *_2, \ldots, *_n}$ be algebraic structures.
Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping.
Then:
:$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *... | As $\paren {\phi^{-1} }^{-1} = \phi$, it suffices to show the sufficient condition.
Suppose that $\phi$ is an isomorphism.
We shall show that $\phi^{-1}$ is also an isomorphism.
We have {{afortiori}} that $\phi$ is a bijection.
Hence from Inverse of Bijection is Bijection we have that $\phi^{-1}$ is also a bijection.
T... | Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ and $\struct {T, *_1, *_2, \ldots, *_n}$ be [[Definition:Algebraic Structure|algebraic structures]].
Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a [[Definition:Mapping|mapping]].
Then:
:$\phi: \struct ... | As $\paren {\phi^{-1} }^{-1} = \phi$, it suffices to show the [[Definition:Sufficient Condition|sufficient condition]].
Suppose that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
We shall show that $\phi^{-1}$ is also an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
We have {{a... | Inverse of Algebraic Structure Isomorphism is Isomorphism/General Result | https://proofwiki.org/wiki/Inverse_of_Algebraic_Structure_Isomorphism_is_Isomorphism/General_Result | https://proofwiki.org/wiki/Inverse_of_Algebraic_Structure_Isomorphism_is_Isomorphism/General_Result | [
"Inverse of Algebraic Structure Isomorphism is Isomorphism"
] | [
"Definition:Algebraic Structure",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Conditional/Sufficient Condition",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract... |
proofwiki-7090 | Fundamental Theorem of Calculus for Complex Riemann Integrals | Let $\closedint a b$ be a closed real interval.
Let $F, f: \closedint a b \to \C$ be complex functions.
Suppose that $F$ is a primitive of $f$.
Then the complex Riemann integral of $f$ satisfies:
:$\ds \int_a^b \map f t \rd t = \map F b - \map F a$ | {{proofread|Missing proof that the conditions for Cauchy-Riemann Equations are satisfied}}
Let $u, v: \closedint a b \times \set 0 \to \R$ be defined as in the Cauchy-Riemann Equations:
:$\map u {t, y} = \map \Re {\map F z}$
:$\map v {t, y} = \map \Im {\map F z}$
where:
:$\map \Re {\map F z}$ denotes the real part of $... | Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
Let $F, f: \closedint a b \to \C$ be [[Definition:Complex Function|complex functions]].
Suppose that $F$ is a [[Definition:Complex Primitive|primitive]] of $f$.
Then the [[Definition:Complex Riemann Integral|complex Riemann integra... | {{proofread|Missing proof that the conditions for Cauchy-Riemann Equations are satisfied}}
Let $u, v: \closedint a b \times \set 0 \to \R$ be defined as in the [[Cauchy-Riemann Equations]]:
:$\map u {t, y} = \map \Re {\map F z}$
:$\map v {t, y} = \map \Im {\map F z}$
where:
:$\map \Re {\map F z}$ denotes the [[Defi... | Fundamental Theorem of Calculus for Complex Riemann Integrals | https://proofwiki.org/wiki/Fundamental_Theorem_of_Calculus_for_Complex_Riemann_Integrals | https://proofwiki.org/wiki/Fundamental_Theorem_of_Calculus_for_Complex_Riemann_Integrals | [
"Complex Analysis",
"Fundamental Theorem of Calculus"
] | [
"Definition:Real Interval/Closed",
"Definition:Complex Function",
"Definition:Primitive (Calculus)/Complex",
"Definition:Integrable Function/Complex"
] | [
"Cauchy-Riemann Equations",
"Definition:Complex Number/Real Part",
"Definition:Complex Number/Imaginary Part",
"Cauchy-Riemann Equations",
"Fundamental Theorem of Calculus"
] |
proofwiki-7091 | Fundamental Theorem of Calculus for Contour Integrals | Let $F, f: D \to \C$ be complex functions, where $D$ is a connected domain.
Let $C$ be a contour that is a concatenation of the directed smooth curves $C_1, \ldots, C_n$.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to D$ for all $k \in \set {1, \ldots, n}$.
Suppose that $F$ is an an... | {{begin-eqn}}
{{eqn | l = \int_C \map f z
| r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t
| c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \paren {\dfrac \rd {\rd t} \map F {\map {\gamma_k} t} } \rd t
| c ... | Let $F, f: D \to \C$ be [[Definition:Complex Function|complex functions]], where $D$ is a [[Definition:Connected Domain (Complex Analysis)|connected domain]].
Let $C$ be a [[Definition:Contour (Complex Plane)|contour]] that is a [[Definition:Concatenation of Contours (Complex Plane)|concatenation]] of the [[Definition... | {{begin-eqn}}
{{eqn | l = \int_C \map f z
| r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \map f {\map {\gamma_k} t} \map {\gamma_k'} t \rd t
| c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \int_{a_k}^{b_k} \paren {\dfrac \rd {\rd t} \map F {\map {\gamma_k} t} } \rd t
| c ... | Fundamental Theorem of Calculus for Contour Integrals | https://proofwiki.org/wiki/Fundamental_Theorem_of_Calculus_for_Contour_Integrals | https://proofwiki.org/wiki/Fundamental_Theorem_of_Calculus_for_Contour_Integrals | [
"Fundamental Theorem of Calculus for Contour Integrals",
"Fundamental Theorem of Calculus",
"Complex Contour Integrals",
"Fundamental Theorems"
] | [
"Definition:Complex Function",
"Definition:Connected Domain (Complex Analysis)",
"Definition:Contour/Complex Plane",
"Definition:Concatenation of Contours/Complex Plane",
"Definition:Directed Smooth Curve/Complex Plane",
"Definition:Directed Smooth Curve/Parameterization/Complex Plane",
"Definition:Smoo... | [
"Derivative of Complex Composite Function",
"Fundamental Theorem of Calculus for Complex Riemann Integrals",
"Definition:Telescoping Series",
"Definition:Contour/Closed/Complex Plane"
] |
proofwiki-7092 | Power Series is Taylor Series | Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a complex power series about $\xi \in \C$.
Let $R$ be the radius of convergence of $f$.
Then, $f$ is of differentiability class $C^\infty$.
For all $n \in \N$:
:$a_n = \dfrac {\map {f^{\paren n} } \xi} {n!}$
Hence, $f$ is equal to its Taylor ser... | First, we prove by induction over $k \in \N_{\ge 1}$ that:
:$\ds \map {f^{\paren k} } z = \sum_{n \mathop = k}^\infty a_n \paren {z - \xi}^{n - k} n^{\underline k}$
where $n^{\underline k}$ denotes the falling factorial. | Let $\ds \map f z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a [[Definition:Complex Power Series|complex power series]] about $\xi \in \C$.
Let $R$ be the [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] of $f$.
Then, $f$ is of [[Definition:Differentiability Class|diff... | First, we prove by [[Principle of Mathematical Induction|induction]] over $k \in \N_{\ge 1}$ that:
:$\ds \map {f^{\paren k} } z = \sum_{n \mathop = k}^\infty a_n \paren {z - \xi}^{n - k} n^{\underline k}$
where $n^{\underline k}$ denotes the [[Definition:Falling Factorial|falling factorial]]. | Power Series is Taylor Series | https://proofwiki.org/wiki/Power_Series_is_Taylor_Series | https://proofwiki.org/wiki/Power_Series_is_Taylor_Series | [
"Complex Power Series",
"Taylor Series"
] | [
"Definition:Power Series/Complex Domain",
"Definition:Radius of Convergence/Complex Domain",
"Definition:Differentiability Class",
"Definition:Taylor Series"
] | [
"Principle of Mathematical Induction",
"Definition:Falling Factorial"
] |
proofwiki-7093 | Image of Complex Exponential Function | The image of the complex exponential function is $\C \setminus \set 0$. | Let $z \in \C \setminus \set 0$.
Let $r = \cmod z$ denote the modulus of $z$.
Let $\theta = \map \arg z$ denote the argument of $z$.
Then $r > 0$.
Let $\ln$ denote the real natural logarithm.
Let $e$ denote the real exponential function.
Then:
{{begin-eqn}}
{{eqn | l = \map \exp {\ln r + i \theta}
| r = e^{\ln r}... | The [[Definition:Image of Mapping|image]] of the [[Definition:Complex Exponential Function|complex exponential function]] is $\C \setminus \set 0$. | Let $z \in \C \setminus \set 0$.
Let $r = \cmod z$ denote the [[Definition:Complex Modulus|modulus]] of $z$.
Let $\theta = \map \arg z$ denote the [[Definition:Argument of Complex Number|argument]] of $z$.
Then $r > 0$.
Let $\ln$ denote the [[Definition:Natural Logarithm|real natural logarithm]].
Let $e$ denote t... | Image of Complex Exponential Function | https://proofwiki.org/wiki/Image_of_Complex_Exponential_Function | https://proofwiki.org/wiki/Image_of_Complex_Exponential_Function | [
"Exponential Function"
] | [
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Exponential Function/Complex"
] | [
"Definition:Complex Modulus",
"Definition:Argument of Complex Number",
"Definition:Natural Logarithm",
"Definition:Exponential Function/Real",
"Exponential of Natural Logarithm",
"Exponential Tends to Zero and Infinity"
] |
proofwiki-7094 | Reciprocal of Complex Exponential | :$\dfrac 1 {\map \exp z} = \map \exp {-z}$ | {{begin-eqn}}
{{eqn | l = \map \exp {-z}
| r = \dfrac {\map \exp {-z} } {\map \exp 0}
| c = as $\map \exp 0 = 1$ by Exponential of Zero
}}
{{eqn | r = \dfrac {\map \exp {-z} } {\map \exp {z - z} }
}}
{{eqn | r = \dfrac {\map \exp {-z} } {\map \exp z \, \map \exp {-z} }
| c = Exponential of Sum: Comple... | :$\dfrac 1 {\map \exp z} = \map \exp {-z}$ | {{begin-eqn}}
{{eqn | l = \map \exp {-z}
| r = \dfrac {\map \exp {-z} } {\map \exp 0}
| c = as $\map \exp 0 = 1$ by [[Exponential of Zero]]
}}
{{eqn | r = \dfrac {\map \exp {-z} } {\map \exp {z - z} }
}}
{{eqn | r = \dfrac {\map \exp {-z} } {\map \exp z \, \map \exp {-z} }
| c = [[Exponential of Sum/C... | Reciprocal of Complex Exponential | https://proofwiki.org/wiki/Reciprocal_of_Complex_Exponential | https://proofwiki.org/wiki/Reciprocal_of_Complex_Exponential | [
"Exponential Function",
"Reciprocals"
] | [] | [
"Exponential of Zero",
"Exponential of Sum/Complex Numbers"
] |
proofwiki-7095 | Power Function Preserves Ordering in Ordered Semigroup | Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y \in S$ such that $x \preceq y$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then:
:$x^n \preceq y^n$
where $x^n$ is the $n$th power of $x$. | By definition of ordered semigroup:
:$\preceq$ is compatible with $\circ$.
By definition of ordering:
:$\preceq$ is transitive.
Thus by Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements:
:$x^n \preceq y^n$
{{qed}} | Let $\struct {S, \circ, \preceq}$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
Let $x, y \in S$ such that $x \preceq y$.
Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Then:
:$x^n \preceq y^n$
where $x^n$ is the $n$th [[Definition:Power of Element of Semig... | By definition of [[Definition:Ordered Semigroup|ordered semigroup]]:
:$\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
By definition of [[Definition:Ordering|ordering]]:
:$\preceq$ is [[Definition:Transitive Relation|transitive]].
Thus by [[Transitive Relation Compatible with Se... | Power Function Preserves Ordering in Ordered Semigroup/Proof 1 | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Semigroup | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Semigroup/Proof_1 | [
"Ordered Semigroups",
"Power Function Preserves Ordering in Ordered Semigroup"
] | [
"Definition:Ordered Semigroup",
"Definition:Strictly Positive/Integer",
"Definition:Power of Element/Semigroup"
] | [
"Definition:Ordered Semigroup",
"Definition:Relation Compatible with Operation",
"Definition:Ordering",
"Definition:Transitive Relation",
"Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements"
] |
proofwiki-7096 | Power Function Preserves Ordering in Ordered Semigroup | Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $x, y \in S$ such that $x \preceq y$.
Let $n \in \N_{>0}$ be a strictly positive integer.
Then:
:$x^n \preceq y^n$
where $x^n$ is the $n$th power of $x$. | The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$x \preceq y \implies x^n \preceq y^n$
$\map P 1$ is the case:
:$x \preceq y \implies x \preceq y$
which is trivially true.
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
We have:
{{begin-eqn}}
{{eqn | l = x
... | Let $\struct {S, \circ, \preceq}$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
Let $x, y \in S$ such that $x \preceq y$.
Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Then:
:$x^n \preceq y^n$
where $x^n$ is the $n$th [[Definition:Power of Element of Semig... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$x \preceq y \implies x^n \preceq y^n$
$\map P 1$ is the case:
:$x \preceq y \implies x \preceq y$
which is trivially true.
Thus $\map P 1$ is seen to ho... | Power Function Preserves Ordering in Ordered Semigroup/Proof 2 | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Semigroup | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Semigroup/Proof_2 | [
"Ordered Semigroups",
"Power Function Preserves Ordering in Ordered Semigroup"
] | [
"Definition:Ordered Semigroup",
"Definition:Strictly Positive/Integer",
"Definition:Power of Element/Semigroup"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Axiom:Ordered Semigroup Axioms",
"Definition:Relation Compatible with Operation",
"Axiom:Ordered Semigroup Axioms",
"Definition:Relation Compatible with Operation",
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:... |
proofwiki-7097 | Power Function Preserves Ordering in Ordered Group | Let $n \in \N_{>0}$ be a strictly positive integer.
Let $\prec$ be the reflexive reduction of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r = x^n \preccurlyeq y^n
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
... | By Power Function Preserves Ordering in Ordered Group:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = x \preccurlyeq e
| o = \implies
| r = x^n \preccurlyeq e^n
}}
{{eqn | l = e \preccurlyeq x
| o = \implies
| r = e^n \preccurlyeq x^n
}}
{{eqn | l = x \prec e
| o = \implies
| ... | Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r ... | By [[Power Function Preserves Ordering in Ordered Group]]:
{{begin-eqn}}
{{eqn | q = \forall x \in S
| l = x \preccurlyeq e
| o = \implies
| r = x^n \preccurlyeq e^n
}}
{{eqn | l = e \preccurlyeq x
| o = \implies
| r = e^n \preccurlyeq x^n
}}
{{eqn | l = x \prec e
| o = \implies
... | Power Function Preserves Ordering in Ordered Group/Corollary/Proof 1 | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group/Corollary/Proof_1 | [
"Ordered Groups",
"Power Function Preserves Ordering in Ordered Group"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Reflexive Reduction",
"Definition:Power of Element/Semigroup"
] | [
"Power Function Preserves Ordering in Ordered Group",
"Identity Element is Idempotent",
"Definition:Idempotence/Element",
"Definition:Idempotence/Element"
] |
proofwiki-7098 | Power Function Preserves Ordering in Ordered Group | Let $n \in \N_{>0}$ be a strictly positive integer.
Let $\prec$ be the reflexive reduction of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r = x^n \preccurlyeq y^n
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
... | By the definition of an ordered group, $\preccurlyeq$ is a transitive relation compatible with $\circ$.
By Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements:
:$x \preccurlyeq e \implies x^n \preccurlyeq e^n$
:$e \preccurlyeq x \implies e^n \preccurlyeq x^n$
By Identity Element i... | Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r ... | By the definition of an [[Definition:Ordered Group|ordered group]], $\preccurlyeq$ is a [[Definition:Transitive Relation|transitive relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
By [[Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements]]:
:$... | Power Function Preserves Ordering in Ordered Group/Corollary/Proof 2 | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group/Corollary/Proof_2 | [
"Ordered Groups",
"Power Function Preserves Ordering in Ordered Group"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Reflexive Reduction",
"Definition:Power of Element/Semigroup"
] | [
"Definition:Ordered Group",
"Definition:Transitive Relation",
"Definition:Relation Compatible with Operation",
"Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements",
"Identity Element is Idempotent",
"Definition:Idempotence/Element",
"Reflexive Reduction of Relatio... |
proofwiki-7099 | Power Function Preserves Ordering in Ordered Group | Let $n \in \N_{>0}$ be a strictly positive integer.
Let $\prec$ be the reflexive reduction of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r = x^n \preccurlyeq y^n
}}
{{eqn | q = \forall x, y \in S
| l = x \prec y
... | By definition of ordered group:
:$\preccurlyeq$ is compatible with $\circ$.
By definition of ordering:
:$\preccurlyeq$ is transitive.
From Reflexive Reduction of Relation Compatible with Group Operation is Compatible:
:$\prec$ is also compatible with $\circ$.
From Reflexive Reduction of Transitive Antisymmetric Relatio... | Let $n \in \N_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preccurlyeq$.
Then the following hold:
{{begin-eqn}}
{{eqn | q = \forall x, y \in S
| l = x \preccurlyeq y
| o = \implies
| r ... | By definition of [[Definition:Ordered Group|ordered group]]:
:$\preccurlyeq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
By definition of [[Definition:Ordering|ordering]]:
:$\preccurlyeq$ is [[Definition:Transitive Relation|transitive]].
From [[Reflexive Reduction of Relation Compati... | Power Function Preserves Ordering in Ordered Group/Proof 1 | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group | https://proofwiki.org/wiki/Power_Function_Preserves_Ordering_in_Ordered_Group/Proof_1 | [
"Ordered Groups",
"Power Function Preserves Ordering in Ordered Group"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Reflexive Reduction",
"Definition:Power of Element/Semigroup"
] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Definition:Ordering",
"Definition:Transitive Relation",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Definition:Relation Compatible with Operation",
"Reflexive Reduction of Transitive Antis... |
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