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-7300 | Factor Principles/Disjunction on Right/Formulation 1 | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
:$\begin{array}{|ccc||ccc||ccccccc|} \hline
p & q & r & (p & \implies & q) & (p & \lor & r) & \implies & (q & \lor & r) \\
\hline
F & F & F ... | :$p \implies q \vdash \paren {p \lor r} \implies \paren {q \lor r}$ | As can be seen for all [[Definition:Boolean Interpretation|boolean interpretations]] by inspection, where the [[Definition:Truth Value|truth value]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] on the {{LHS}} is $T$, that under the one on the {{RHS}} is also $T$:
:$\begin{array}{|ccc... | Factor Principles/Disjunction on Right/Formulation 1/Proof 4 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Right/Formulation_1/Proof_4 | [
"Factor Principles"
] | [] | [
"Definition:Boolean Interpretation",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic"
] |
proofwiki-7301 | Factor Principles/Disjunction on Left/Formulation 1 | : $p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$ | {{BeginTableau|p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q} }}
{{Premise|1|p \implies q}}
{{TheoremIntro|2|r \implies r|Law of Identity: Formulation 2}}
{{SequentIntro|3|1|\paren {r \lor p} \implies \paren {r \lor q}|2, 1|Constructive Dilemma}}
{{EndTableau}}
{{qed}} | : $p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$ | {{BeginTableau|p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q} }}
{{Premise|1|p \implies q}}
{{TheoremIntro|2|r \implies r|[[Law of Identity/Formulation 2|Law of Identity: Formulation 2]]}}
{{SequentIntro|3|1|\paren {r \lor p} \implies \paren {r \lor q}|2, 1|[[Constructive Dilemma/Formulation 1|Constru... | Factor Principles/Disjunction on Left/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_1/Proof_1 | [
"Factor Principles"
] | [] | [
"Law of Identity/Formulation 2",
"Constructive Dilemma/Formulation 1"
] |
proofwiki-7302 | Factor Principles/Disjunction on Left/Formulation 1 | : $p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$ | {{BeginTableau|p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q} }}
{{Premise|1|p \implies q}}
{{Assumption|2|r \lor p}}
{{Assumption|3|r}}
{{Addition|4|3|r \lor q|3|1}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|1|5}}
{{Addition|7|1, 5|r \lor q|6|2}}
{{ProofByCases|8|1, 2|r \lor q|2|3|4|5|6}}
{{Implicati... | : $p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q}$ | {{BeginTableau|p \implies q \vdash \paren {r \lor p} \implies \paren {r \lor q} }}
{{Premise|1|p \implies q}}
{{Assumption|2|r \lor p}}
{{Assumption|3|r}}
{{Addition|4|3|r \lor q|3|1}}
{{Assumption|5|p}}
{{ModusPonens|6|1, 5|q|1|5}}
{{Addition|7|1, 5|r \lor q|6|2}}
{{ProofByCases|8|1, 2|r \lor q|2|3|4|5|6}}
{{Implicati... | Factor Principles/Disjunction on Left/Formulation 1/Proof 2 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_1 | https://proofwiki.org/wiki/Factor_Principles/Disjunction_on_Left/Formulation_1/Proof_2 | [
"Factor Principles"
] | [] | [] |
proofwiki-7303 | Disjunction of Conditionals | :$\vdash \paren {p \implies q} \lor \paren {q \implies r}$ | {{BeginTableau|\vdash \paren {p \implies q} \lor \paren {q \implies r} }}
{{ExcludedMiddle|1|q \lor \neg q}}
{{Assumption|2|q}}
{{SequentIntro|3|2|p \implies q|2|True Statement is implied by Every Statement}}
{{Addition|4|2|\paren {p \implies q} \lor \paren {q \implies r}|3|1}}
{{Assumption|5|\neg q}}
{{SequentIntro|6|... | :$\vdash \paren {p \implies q} \lor \paren {q \implies r}$ | {{BeginTableau|\vdash \paren {p \implies q} \lor \paren {q \implies r} }}
{{ExcludedMiddle|1|q \lor \neg q}}
{{Assumption|2|q}}
{{SequentIntro|3|2|p \implies q|2|[[True Statement is implied by Every Statement]]}}
{{Addition|4|2|\paren {p \implies q} \lor \paren {q \implies r}|3|1}}
{{Assumption|5|\neg q}}
{{SequentIntr... | Disjunction of Conditionals | https://proofwiki.org/wiki/Disjunction_of_Conditionals | https://proofwiki.org/wiki/Disjunction_of_Conditionals | [
"Disjunction",
"Conditional"
] | [] | [
"True Statement is implied by Every Statement",
"False Statement implies Every Statement"
] |
proofwiki-7304 | Principle of Composition/Formulation 1/Forward Implication | :$\paren {p \implies r} \lor \paren {q \implies r} \vdash \paren {p \land q} \implies r$ | {{BeginTableau|\paren {p \implies r} \lor \paren {q \implies r} \vdash \paren {p \land q} \implies r}}
{{Premise | 1 | \paren {p \implies r} \lor \paren {q \implies r} }}
{{Assumption | 2 | p \implies r}}
{{Assumption | 3 | p \land q}}
{{Simplification | 4 | 3 | p | 3 | 1}}
{{ModusPonens | 5 | 2,... | :$\paren {p \implies r} \lor \paren {q \implies r} \vdash \paren {p \land q} \implies r$ | {{BeginTableau|\paren {p \implies r} \lor \paren {q \implies r} \vdash \paren {p \land q} \implies r}}
{{Premise | 1 | \paren {p \implies r} \lor \paren {q \implies r} }}
{{Assumption | 2 | p \implies r}}
{{Assumption | 3 | p \land q}}
{{Simplification | 4 | 3 | p | 3 | 1}}
{{ModusPonens | 5 | 2,... | Principle of Composition/Formulation 1/Forward Implication | https://proofwiki.org/wiki/Principle_of_Composition/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Principle_of_Composition/Formulation_1/Forward_Implication | [
"Principle of Composition"
] | [] | [
"Category:Principle of Composition"
] |
proofwiki-7305 | Principle of Composition/Formulation 1/Reverse Implication | :$\paren {p \land q} \implies r \vdash \paren {p \implies r} \lor \paren {q \implies r}$ | {{BeginTableau|\paren {p \land q} \implies r \vdash \paren {p \implies r} \lor \paren {q \implies r} }}
{{Premise | 1 | \paren {p \land q} \implies r}}
{{SequentIntro | 2 | 1 | \neg \paren {p \lor q} \lor r | 1 | Rule of Material Implication}}
{{SequentIntro | 3 | 1 | \neg p \lor \neg q \lor r | 2 | De Mo... | :$\paren {p \land q} \implies r \vdash \paren {p \implies r} \lor \paren {q \implies r}$ | {{BeginTableau|\paren {p \land q} \implies r \vdash \paren {p \implies r} \lor \paren {q \implies r} }}
{{Premise | 1 | \paren {p \land q} \implies r}}
{{SequentIntro | 2 | 1 | \neg \paren {p \lor q} \lor r | 1 | [[Rule of Material Implication/Formulation 1|Rule of Material Implication]]}}
{{SequentIntro |... | Principle of Composition/Formulation 1/Reverse Implication | https://proofwiki.org/wiki/Principle_of_Composition/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Principle_of_Composition/Formulation_1/Reverse_Implication | [
"Principle of Composition"
] | [] | [
"Rule of Material Implication/Formulation 1",
"De Morgan's Laws (Logic)/Disjunction of Negations",
"Rule of Material Implication/Formulation 1",
"Rule of Material Implication/Formulation 1",
"Category:Principle of Composition"
] |
proofwiki-7306 | Principle of Composition/Formulation 2 | :$\paren {\paren {p \implies r} \lor \paren {q \implies r} } \iff \paren {\paren {p \land q} \implies r}$ | {{BeginTableau|\paren {\paren {p \implies r} \lor \paren {q \implies r} } \iff \paren {\paren {p \land q} \implies r} }}
{{Assumption |1|\paren {p \implies r} \lor \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \land q} \implies r|1|Principle of Composition: Formulation 1}}
{{Implication |3||\paren {\paren {p \... | :$\paren {\paren {p \implies r} \lor \paren {q \implies r} } \iff \paren {\paren {p \land q} \implies r}$ | {{BeginTableau|\paren {\paren {p \implies r} \lor \paren {q \implies r} } \iff \paren {\paren {p \land q} \implies r} }}
{{Assumption |1|\paren {p \implies r} \lor \paren {q \implies r} }}
{{SequentIntro|2|1|\paren {p \land q} \implies r|1|[[Principle of Composition/Formulation 1/Forward Implication|Principle of Compo... | Principle of Composition/Formulation 2 | https://proofwiki.org/wiki/Principle_of_Composition/Formulation_2 | https://proofwiki.org/wiki/Principle_of_Composition/Formulation_2 | [
"Principle of Composition"
] | [] | [
"Principle of Composition/Formulation 1/Forward Implication",
"Principle of Composition/Formulation 1/Reverse Implication"
] |
proofwiki-7307 | Inversion Mapping on Topological Group is Homeomorphism | Let $T = \struct {G, \circ, \tau}$ be a topological group.
Let $\phi: G \to G$ be the inversion mapping of $T$.
Then $\phi$ is a homeomorphism. | From the definition of topological group, $\phi$ is continuous.
By Inversion Mapping is Involution, $\phi$ is an involution.
By Continuous Involution is Homeomorphism, $\phi$ is a homeomorphism.
{{qed}}
Category:Inversion Mappings
Category:Homeomorphisms (Topological Spaces)
Category:Topological Groups
dd53003fsqyn6lrl... | Let $T = \struct {G, \circ, \tau}$ be a [[Definition:Topological Group|topological group]].
Let $\phi: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] of $T$.
Then $\phi$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]]. | From the definition of [[Definition:Topological Group|topological group]], $\phi$ is [[Definition:Continuous Mapping (Topology)|continuous]].
By [[Inversion Mapping is Involution]], $\phi$ is an [[Definition:Involution (Mapping)|involution]].
By [[Continuous Involution is Homeomorphism]], $\phi$ is a [[Definition:Hom... | Inversion Mapping on Topological Group is Homeomorphism | https://proofwiki.org/wiki/Inversion_Mapping_on_Topological_Group_is_Homeomorphism | https://proofwiki.org/wiki/Inversion_Mapping_on_Topological_Group_is_Homeomorphism | [
"Inversion Mappings",
"Homeomorphisms (Topological Spaces)",
"Topological Groups"
] | [
"Definition:Topological Group",
"Definition:Inversion Mapping",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Definition:Topological Group",
"Definition:Continuous Mapping (Topology)",
"Inversion Mapping is Involution",
"Definition:Involution (Mapping)",
"Continuous Involution is Homeomorphism",
"Definition:Homeomorphism/Topological Spaces",
"Category:Inversion Mappings",
"Category:Homeomorphisms (Topologica... |
proofwiki-7308 | De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Forward Implication | :$\paren {\neg p \land \neg q} \implies \paren {\neg \paren {p \lor q} }$ | {{BeginTableau|\paren {\neg p \land \neg q} \implies \paren {\neg \paren {p \lor q} } }}
{{Assumption|1|\neg p \land \neg q}}
{{SequentIntro|2|1|\neg \paren {p \lor q}|1|De Morgan's Laws (Logic): Conjunction of Negations: Formulation 1}}
{{Implication|3||\paren {\neg p \land \neg q} \implies \paren {\neg \paren {p \lor... | :$\paren {\neg p \land \neg q} \implies \paren {\neg \paren {p \lor q} }$ | {{BeginTableau|\paren {\neg p \land \neg q} \implies \paren {\neg \paren {p \lor q} } }}
{{Assumption|1|\neg p \land \neg q}}
{{SequentIntro|2|1|\neg \paren {p \lor q}|1|[[De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication|De Morgan's Laws (Logic): Conjunction of Negations: Formulation ... | De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Forward Implication | https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2/Forward_Implication | [
"De Morgan's Laws (Logic)"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Forward Implication"
] |
proofwiki-7309 | De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Reverse Implication | :$\paren {\neg \paren {p \lor q} } \implies \paren {\neg p \land \neg q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \lor q} } \implies \paren {\neg p \land \neg q} }}
{{Assumption|1|\neg \paren {p \lor q} }}
{{SequentIntro|2|1|\neg p \land \neg q|1|De Morgan's Laws (Logic): Conjunction of Negations: Formulation 1}}
{{Implication|3||\paren {\neg \paren {p \lor q} } \implies \paren {\neg p ... | :$\paren {\neg \paren {p \lor q} } \implies \paren {\neg p \land \neg q}$ | {{BeginTableau|\vdash \paren {\neg \paren {p \lor q} } \implies \paren {\neg p \land \neg q} }}
{{Assumption|1|\neg \paren {p \lor q} }}
{{SequentIntro|2|1|\neg p \land \neg q|1|[[De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication|De Morgan's Laws (Logic): Conjunction of Negations: Form... | De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 2/Reverse Implication | https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/De_Morgan's_Laws_(Logic)/Conjunction_of_Negations/Formulation_2/Reverse_Implication | [
"De Morgan's Laws (Logic)"
] | [] | [
"De Morgan's Laws (Logic)/Conjunction of Negations/Formulation 1/Reverse Implication",
"Category:De Morgan's Laws (Logic)"
] |
proofwiki-7310 | Equivalence of Definitions of Generalized Ordered Space | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\tau$ be a topology for $S$.
{{TFAE|def = Generalized Ordered Space}} | === Definition $(1)$ implies Definition $(3)$ ===
{{:Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3}}{{qed|lemma}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\tau$ be a [[Definition:Topology|topology]] for $S$.
{{TFAE|def = Generalized Ordered Space}} | === [[Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3|Definition $(1)$ implies Definition $(3)$]] ===
{{:Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3}}{{qed|lemma}} | Equivalence of Definitions of Generalized Ordered Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space | [
"Generalized Ordered Spaces",
"Equivalence of Definitions of Generalized Ordered Space"
] | [
"Definition:Totally Ordered Set",
"Definition:Topology"
] | [
"Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3"
] |
proofwiki-7311 | Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1 | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 2:
{{:Definition:Generalized Ordered Space/Definition 2}}
Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 1:
{{:Definition:Generalized Ordered Space/Definition 1}} | Let $x \in U \in \tau$.
Then by the definition of topological embedding:
:$\map \phi U$ is an open neighborhood of $\map \phi x$ in $\map \phi S$ with the subspace topology.
Thus by Basis for Topological Subspace and the definition of the order topology, there is an open interval or open ray $I' \in \tau'$ such that:
:... | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space/Definition 2|generalized ordered space by Definition 2]]:
{{:Definition:Generalized Ordered Space/Definition 2}}
Then $\struct {S, \preceq, \tau}$ is a [[Definition:Generalized Ordered Space/Definition 1|generalized ordered space by Definitio... | Let $x \in U \in \tau$.
Then by the definition of [[Definition:Topological Embedding|topological embedding]]:
:$\map \phi U$ is an [[Definition:Open Neighborhood of Point|open neighborhood]] of $\map \phi x$ in $\map \phi S$ with the [[Definition:Subspace Topology|subspace topology]].
Thus by [[Basis for Topological ... | Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space/Definition_2_implies_Definition_1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space/Definition_2_implies_Definition_1 | [
"Equivalence of Definitions of Generalized Ordered Space"
] | [
"Definition:Generalized Ordered Space/Definition 2",
"Definition:Generalized Ordered Space/Definition 1"
] | [
"Definition:Embedding (Topology)",
"Definition:Open Neighborhood/Point",
"Definition:Topological Subspace",
"Basis for Topological Subspace",
"Definition:Order Topology",
"Definition:Interval/Ordered Set/Open",
"Definition:Ray (Order Theory)/Open",
"Definition:Interval/Ordered Set",
"Definition:Ray ... |
proofwiki-7312 | Inverse Image of Order-Convex Set under Monotone Mapping is Order-Convex | Let $\struct {X, \le}$ and $\struct {Y, \preceq}$ be ordered sets.
Let $f: X \to Y$ be a monotone mapping.
Let $C$ be a order-convex subset of $Y$.
Then $f^{-1} \sqbrk C$ is order-convex in $X$. | Suppose $f$ is increasing.
Let $a, b, c \in X$ such that $a \le b \le c$.
Let $a, c \in f^{-1} \sqbrk C$.
By definition of inverse image:
:$\map f a, \map f c \in C$
By definition of increasing mapping:
:$\map f a \preceq \map f b \preceq \map f c$
Thus by definition of order-convex set:
:$\map f b \in C$
Then by defin... | Let $\struct {X, \le}$ and $\struct {Y, \preceq}$ be [[Definition:Ordered Set|ordered sets]].
Let $f: X \to Y$ be a [[Definition:Monotone Mapping|monotone mapping]].
Let $C$ be a [[Definition:Order-Convex Set|order-convex]] [[Definition:Subset|subset]] of $Y$.
Then $f^{-1} \sqbrk C$ is [[Definition:Order-Convex Set... | Suppose $f$ is [[Definition:Increasing Mapping|increasing]].
Let $a, b, c \in X$ such that $a \le b \le c$.
Let $a, c \in f^{-1} \sqbrk C$.
By definition of [[Definition:Inverse Image|inverse image]]:
:$\map f a, \map f c \in C$
By definition of [[Definition:Increasing Mapping|increasing mapping]]:
:$\map f a \pre... | Inverse Image of Order-Convex Set under Monotone Mapping is Order-Convex | https://proofwiki.org/wiki/Inverse_Image_of_Order-Convex_Set_under_Monotone_Mapping_is_Order-Convex | https://proofwiki.org/wiki/Inverse_Image_of_Order-Convex_Set_under_Monotone_Mapping_is_Order-Convex | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Monotone (Order Theory)/Mapping",
"Definition:Order-Convex Set",
"Definition:Subset",
"Definition:Order-Convex Set"
] | [
"Definition:Increasing/Mapping",
"Definition:Inverse Image",
"Definition:Increasing/Mapping",
"Definition:Order-Convex Set",
"Definition:Inverse Image",
"Definition:Order-Convex Set",
"Definition:Decreasing/Mapping",
"Category:Order Theory"
] |
proofwiki-7313 | Conjunction with Law of Excluded Middle | :$\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q}$ | {{BeginTableau|\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q} }}
{{Assumption|1|p}}
{{ExcludedMiddle|2|q \lor \neg q}}
{{Assumption|3|q}}
{{Conjunction|4|1, 3|p \land q|1|2}}
{{Addition|5|1, 3|\paren {p \land q} \lor \paren {p \land \neg q}|4|1}}
{{Assumption|6|\neg q}}
{{Conjunction|7|1, 6|p \land \neg ... | :$\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q}$ | {{BeginTableau|\vdash p \iff \paren {p \land q} \lor \paren {p \land \neg q} }}
{{Assumption|1|p}}
{{ExcludedMiddle|2|q \lor \neg q}}
{{Assumption|3|q}}
{{Conjunction|4|1, 3|p \land q|1|2}}
{{Addition|5|1, 3|\paren {p \land q} \lor \paren {p \land \neg q}|4|1}}
{{Assumption|6|\neg q}}
{{Conjunction|7|1, 6|p \land \neg ... | Conjunction with Law of Excluded Middle | https://proofwiki.org/wiki/Conjunction_with_Law_of_Excluded_Middle | https://proofwiki.org/wiki/Conjunction_with_Law_of_Excluded_Middle | [
"Conjunction",
"Law of Excluded Middle"
] | [] | [
"Rule of Distribution/Conjunction Distributes over Disjunction"
] |
proofwiki-7314 | Proof by Cases with Contradiction | :$\vdash p \iff \paren {p \lor q} \land \paren {p \lor \neg q}$ | {{BeginTableau|\vdash p \iff \paren {p \lor q} \land \paren {p \lor \neg q} }}
{{Assumption|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{Addition|3|1|p \lor \neg q|1|2}}
{{Conjunction|4|1|\paren {p \lor q} \land \paren {p \lor \neg q} |2|3}}
{{Implication|5||p \implies \paren {p \lor q} \land \paren {p \lor \neg q}|1|4}}
{{As... | :$\vdash p \iff \paren {p \lor q} \land \paren {p \lor \neg q}$ | {{BeginTableau|\vdash p \iff \paren {p \lor q} \land \paren {p \lor \neg q} }}
{{Assumption|1|p}}
{{Addition|2|1|p \lor q|1|1}}
{{Addition|3|1|p \lor \neg q|1|2}}
{{Conjunction|4|1|\paren {p \lor q} \land \paren {p \lor \neg q} |2|3}}
{{Implication|5||p \implies \paren {p \lor q} \land \paren {p \lor \neg q}|1|4}}
{{As... | Proof by Cases with Contradiction | https://proofwiki.org/wiki/Proof_by_Cases_with_Contradiction | https://proofwiki.org/wiki/Proof_by_Cases_with_Contradiction | [
"Principle of Non-Contradiction",
"Proof by Cases"
] | [] | [
"Rule of Distribution/Disjunction Distributes over Conjunction",
"Principle of Non-Contradiction/Sequent Form/Formulation 2"
] |
proofwiki-7315 | Ray is Order-Convex | Let $\struct {S, \preceq}$ be an ordered set.
Let $I$ be a ray, either open or closed.
Then $I$ is order-convex in $S$. | The cases for upward-pointing and downward-pointing rays are equivalent.
{{explain|"Dual of convex is convex" and duality, upon which the above statement depends.}}
{{WLOG}}, suppose that $U$ is an upward-pointing ray.
By the definition of a ray, there exists an $a \in S$ such that:
:$I = a^\succ$
or;
:$I = a^\succeq$
... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $I$ be a [[Definition:Ray (Order Theory)|ray]], either [[Definition:Open Ray|open]] or [[Definition:Closed Ray|closed]].
Then $I$ is [[Definition:Order-Convex Set|order-convex]] in $S$. | The cases for [[Definition:Upward-Pointing Ray|upward-pointing]] and [[Definition:Downward-Pointing Ray|downward-pointing]] [[Definition:Ray (Order Theory)|rays]] are equivalent.
{{explain|"Dual of convex is convex" and duality, upon which the above statement depends.}}
{{WLOG}}, suppose that $U$ is an [[Definition:U... | Ray is Order-Convex | https://proofwiki.org/wiki/Ray_is_Order-Convex | https://proofwiki.org/wiki/Ray_is_Order-Convex | [
"Order-Convex Sets",
"Rays (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Ray (Order Theory)",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Order Theory)/Closed",
"Definition:Order-Convex Set"
] | [
"Definition:Ray (Order Theory)/Upward-Pointing",
"Definition:Ray (Order Theory)/Downward-Pointing",
"Definition:Ray (Order Theory)",
"Definition:Ray (Order Theory)/Upward-Pointing",
"Definition:Ray (Order Theory)",
"Definition:Ray (Order Theory)/Open",
"Definition:Ray (Order Theory)/Closed",
"Definiti... |
proofwiki-7316 | Intersection of Order-Convex Sets is Order-Convex | Let $\struct {S, \preceq}$ be an ordered set.
Let $\CC$ be a set of order-convex sets in $S$.
Then $\bigcap \CC$ is order-convex. | Let $a, b, c \in S$.
Let $a, c \in \bigcap \CC$.
Let $a \prec b \prec c$.
By the definition of intersection:
:$\forall T \in \CC: a, c \in T$
Since each $T \in \CC$ is order-convex:
:$\forall T \in \CC: b \in T$
Thus by the definition of intersection:
:$b \in \bigcap \CC$
Thus $\bigcap \CC$ is order-convex.
{{qed}}
Cat... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\CC$ be a [[Definition:Set|set]] of [[Definition:Order-Convex Set|order-convex sets]] in $S$.
Then $\bigcap \CC$ is [[Definition:Order-Convex Set|order-convex]]. | Let $a, b, c \in S$.
Let $a, c \in \bigcap \CC$.
Let $a \prec b \prec c$.
By the definition of [[Definition:Set Intersection|intersection]]:
:$\forall T \in \CC: a, c \in T$
Since each $T \in \CC$ is [[Definition:Order-Convex Set|order-convex]]:
:$\forall T \in \CC: b \in T$
Thus by the definition of [[Definiti... | Intersection of Order-Convex Sets is Order-Convex | https://proofwiki.org/wiki/Intersection_of_Order-Convex_Sets_is_Order-Convex | https://proofwiki.org/wiki/Intersection_of_Order-Convex_Sets_is_Order-Convex | [
"Order-Convex Sets"
] | [
"Definition:Ordered Set",
"Definition:Set",
"Definition:Order-Convex Set",
"Definition:Order-Convex Set"
] | [
"Definition:Set Intersection",
"Definition:Order-Convex Set",
"Definition:Set Intersection",
"Definition:Order-Convex Set",
"Category:Order-Convex Sets"
] |
proofwiki-7317 | Upper and Lower Closures are Order-Convex | Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$.
Then $a^\succeq$, $a^\succ$, $a^\preceq$, and $a^\prec$ are order-convex in $S$. | The cases for upper and lower closures are dual, so we need only prove the case for upper closures.
Suppose, then, that $C = a^\succeq$ or $C = a^\succ$.
Suppose that $x, y, z \in S$, $x \prec y \prec z$, and $x, z \in C$.
Then $a \preceq x \prec y$, so $a \prec y$ by Extended Transitivity.
Therefore $y \in a^\succ \su... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $a \in S$.
Then $a^\succeq$, $a^\succ$, $a^\preceq$, and $a^\prec$ are [[Definition:Order-Convex Set|order-convex]] in $S$. | The cases for upper and lower closures are dual, so we need only prove the case for upper closures.
Suppose, then, that $C = a^\succeq$ or $C = a^\succ$.
Suppose that $x, y, z \in S$, $x \prec y \prec z$, and $x, z \in C$.
Then $a \preceq x \prec y$, so $a \prec y$ by [[Extended Transitivity]].
Therefore $y \in a^\... | Upper and Lower Closures are Order-Convex | https://proofwiki.org/wiki/Upper_and_Lower_Closures_are_Order-Convex | https://proofwiki.org/wiki/Upper_and_Lower_Closures_are_Order-Convex | [
"Lower Closures",
"Upper Closures",
"Order-Convex Sets"
] | [
"Definition:Ordered Set",
"Definition:Order-Convex Set"
] | [
"Extended Transitivity",
"Definition:Order-Convex Set",
"Category:Lower Closures",
"Category:Upper Closures",
"Category:Order-Convex Sets"
] |
proofwiki-7318 | Transitive Closure of Symmetric Relation is Symmetric | Let $S$ be a set.
Let $\RR$ be a symmetric relation on $S$.
Let $\TT$ be the transitive closure of $\RR$.
The $\TT$ is symmetric. | Let $a, b \in S$ with $a \mathrel \TT b$.
By the definition of transitive closure, there is an $n \in \N$ such that $a \mathrel {\RR^n} b $.
Thus there are $x_0, x_1, \dots x_n \in S$ such that:
:$x_0 = a$
:$x_n = b$
:For $k = 0, \dots, n-1$: $x_k \mathrel \RR x_{k + 1}$
For $k = 0, \dots, n$, let $y_k = x_{n - k}$.
Th... | Let $S$ be a [[Definition:set|set]].
Let $\RR$ be a [[Definition:Symmetric Relation|symmetric relation]] on $S$.
Let $\TT$ be the [[Definition:Transitive Closure of Relation|transitive closure]] of $\RR$.
The $\TT$ is [[Definition:Symmetric Relation|symmetric]]. | Let $a, b \in S$ with $a \mathrel \TT b$.
By the definition of [[Definition:Transitive Closure of Relation/Union of Compositions|transitive closure]], there is an $n \in \N$ such that $a \mathrel {\RR^n} b $.
Thus there are $x_0, x_1, \dots x_n \in S$ such that:
:$x_0 = a$
:$x_n = b$
:For $k = 0, \dots, n-1$: $x_k \m... | Transitive Closure of Symmetric Relation is Symmetric | https://proofwiki.org/wiki/Transitive_Closure_of_Symmetric_Relation_is_Symmetric | https://proofwiki.org/wiki/Transitive_Closure_of_Symmetric_Relation_is_Symmetric | [
"Symmetric Relations",
"Transitive Closures"
] | [
"Definition:set",
"Definition:Symmetric Relation",
"Definition:Transitive Closure of Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Transitive Closure of Relation/Union of Compositions",
"Definition:Symmetric Relation",
"Category:Symmetric Relations",
"Category:Transitive Closures"
] |
proofwiki-7319 | Transitive Closure of Reflexive Relation is Reflexive | Let $S$ be a set.
Let $\RR$ be a reflexive relation on $S$.
Let $\RR^+$ be the transitive closure of $\RR$.
Then $\RR^+$ is reflexive. | Let $a \in S$.
Since $\RR$ is reflexive:
:$\tuple {a, a} \in \RR$
By the definition of transitive closure:
:$\RR \subseteq \RR^+$
Thus by the definition of subset:
:$\tuple {a, a} \in \RR^+$
Since this holds for all $a \in S$, it follows that $\RR^+$ is reflexive.
{{qed}} | Let $S$ be a [[Definition:set|set]].
Let $\RR$ be a [[Definition:Reflexive Relation|reflexive relation]] on $S$.
Let $\RR^+$ be the [[Definition:Transitive Closure of Relation|transitive closure]] of $\RR$.
Then $\RR^+$ is [[Definition:Reflexive Relation|reflexive]]. | Let $a \in S$.
Since $\RR$ is [[Definition:Reflexive Relation|reflexive]]:
:$\tuple {a, a} \in \RR$
By the definition of [[Definition:Transitive Closure of Relation/Smallest Transitive Superset|transitive closure]]:
:$\RR \subseteq \RR^+$
Thus by the definition of [[Definition:Subset|subset]]:
:$\tuple {a, a} \in \R... | Transitive Closure of Reflexive Relation is Reflexive | https://proofwiki.org/wiki/Transitive_Closure_of_Reflexive_Relation_is_Reflexive | https://proofwiki.org/wiki/Transitive_Closure_of_Reflexive_Relation_is_Reflexive | [
"Reflexive Relations",
"Transitive Closures"
] | [
"Definition:set",
"Definition:Reflexive Relation",
"Definition:Transitive Closure of Relation",
"Definition:Reflexive Relation"
] | [
"Definition:Reflexive Relation",
"Definition:Transitive Closure of Relation/Smallest Transitive Superset",
"Definition:Subset",
"Definition:Reflexive Relation"
] |
proofwiki-7320 | Transitive Closure of Reflexive Symmetric Relation is Equivalence | Let $S$ be a set.
Let $\RR$ be a symmetric and reflexive relation on $S$.
Then the transitive closure of $\RR$ is an equivalence relation. | Let $\sim$ be the transitive closure of $\RR$.
Checking in turn each of the criteria for equivalence: | Let $S$ be a [[Definition:Set|set]].
Let $\RR$ be a [[Definition:Symmetric Relation|symmetric]] and [[Definition:Reflexive Relation|reflexive]] [[Definition:Endorelation|relation]] on $S$.
Then the [[Definition:Transitive Closure of Relation|transitive closure]] of $\RR$ is an [[Definition:Equivalence Relation|equiv... | Let $\sim$ be the [[Definition:Transitive Closure of Relation|transitive closure]] of $\RR$.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Transitive Closure of Reflexive Symmetric Relation is Equivalence | https://proofwiki.org/wiki/Transitive_Closure_of_Reflexive_Symmetric_Relation_is_Equivalence | https://proofwiki.org/wiki/Transitive_Closure_of_Reflexive_Symmetric_Relation_is_Equivalence | [
"Transitive Closures",
"Equivalence Relations"
] | [
"Definition:Set",
"Definition:Symmetric Relation",
"Definition:Reflexive Relation",
"Definition:Endorelation",
"Definition:Transitive Closure of Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Transitive Closure of Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-7321 | Union of Overlapping Order-Convex Sets in Toset is Order-Convex | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $U$ and $V$ be order-convex sets in $S$.
Let $U \cap V \ne \O$.
Then $U \cup V$ is also order-convex. | Let $a,b,c \in S$
Let $a,c \in U \cup V$.
Let $a \prec b \prec c$.
If $a, c \in U$ then $b \in U$ because $U$ is order-convex.
Thus $b \in U \cup V$ by the definition of union.
Similarly, if $a,c \in V$ then $b \in U \cup V$.
Otherwise, {{WLOG}}, suppose that $a \in U$ and $c \in V$.
Since $U \cap V$ is nonempty by the... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $U$ and $V$ be [[Definition:Order-Convex Set|order-convex sets]] in $S$.
Let $U \cap V \ne \O$.
Then $U \cup V$ is also [[Definition:Order-Convex Set|order-convex]]. | Let $a,b,c \in S$
Let $a,c \in U \cup V$.
Let $a \prec b \prec c$.
If $a, c \in U$ then $b \in U$ because $U$ is [[Definition:Order-Convex Set|order-convex]].
Thus $b \in U \cup V$ by the definition of [[Definition:Set Union|union]].
Similarly, if $a,c \in V$ then $b \in U \cup V$.
Otherwise, {{WLOG}}, suppose th... | Union of Overlapping Order-Convex Sets in Toset is Order-Convex | https://proofwiki.org/wiki/Union_of_Overlapping_Order-Convex_Sets_in_Toset_is_Order-Convex | https://proofwiki.org/wiki/Union_of_Overlapping_Order-Convex_Sets_in_Toset_is_Order-Convex | [
"Order-Convex Sets",
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Order-Convex Set",
"Definition:Order-Convex Set"
] | [
"Definition:Order-Convex Set",
"Definition:Set Union",
"Definition:Total Ordering",
"Definition:Order-Convex Set",
"Definition:Order-Convex Set",
"Category:Order-Convex Sets",
"Category:Total Orderings"
] |
proofwiki-7322 | Rule of Material Equivalence | The '''Rule of Material Equivalence''' is a valid deduction sequent in propositional logic:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==== Formulation 1 ====
{{:Rule of Material Equivalence/Formulation 1}}
==== Formulation 2 ====
{... | {{BeginTableau|p \iff q \vdash \paren {p \implies q} \land \paren {q \implies p} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{Conjunction|4|1|\paren {p \implies q} \land \paren {q \implies p}|2|3}}
{{EndTableau}}
{{BeginTableau|\paren {p... | The '''[[Rule of Material Equivalence]]''' is a [[Definition:Valid Argument|valid]] deduction [[Definition:Sequent|sequent]] in [[Definition:Propositional Logic|propositional logic]]:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==... | {{BeginTableau|p \iff q \vdash \paren {p \implies q} \land \paren {q \implies p} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{Conjunction|4|1|\paren {p \implies q} \land \paren {q \implies p}|2|3}}
{{EndTableau}}
{{BeginTableau|\paren ... | Rule of Material Equivalence/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1/Proof_1 | [
"Rule of Material Equivalence",
"Biconditional",
"Conditional",
"Conjunction"
] | [
"Rule of Material Equivalence",
"Definition:Valid Argument",
"Definition:Sequent",
"Definition:Propositional Logic",
"Rule of Material Equivalence/Formulation 1",
"Rule of Material Equivalence/Formulation 2"
] | [] |
proofwiki-7323 | Rule of Material Equivalence | The '''Rule of Material Equivalence''' is a valid deduction sequent in propositional logic:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==== Formulation 1 ====
{{:Rule of Material Equivalence/Formulation 1}}
==== Formulation 2 ====
{... | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|ccc|ccccccc|} \hline
p & \iff & q & (p & \implies & q) & \land & (q & \implies & p) \\
\hline
\F & \T & \F & \F & \T & \F & \T & \F & \T & \F \\
... | The '''[[Rule of Material Equivalence]]''' is a [[Definition:Valid Argument|valid]] deduction [[Definition:Sequent|sequent]] in [[Definition:Propositional Logic|propositional logic]]:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==... | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|ccc|ccccccc|} \hline
... | Rule of Material Equivalence/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Material_Equivalence | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1/Proof_by_Truth_Table | [
"Rule of Material Equivalence",
"Biconditional",
"Conditional",
"Conjunction"
] | [
"Rule of Material Equivalence",
"Definition:Valid Argument",
"Definition:Sequent",
"Definition:Propositional Logic",
"Rule of Material Equivalence/Formulation 1",
"Rule of Material Equivalence/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7324 | Rule of Material Equivalence | The '''Rule of Material Equivalence''' is a valid deduction sequent in propositional logic:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==== Formulation 1 ====
{{:Rule of Material Equivalence/Formulation 1}}
==== Formulation 2 ====
{... | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} } }}
{{Assumption|1|p \iff q}}
{{SequentIntro|2|1|\paren {\paren {p \implies q} \land \paren {q \implies p} }|1|Rule of Material Equivalence: Formulation 1}}
{{Implication|3||\paren {p \iff q} \implies \paren {\paren ... | The '''[[Rule of Material Equivalence]]''' is a [[Definition:Valid Argument|valid]] deduction [[Definition:Sequent|sequent]] in [[Definition:Propositional Logic|propositional logic]]:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==... | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} } }}
{{Assumption|1|p \iff q}}
{{SequentIntro|2|1|\paren {\paren {p \implies q} \land \paren {q \implies p} }|1|[[Rule of Material Equivalence/Formulation 1|Rule of Material Equivalence: Formulation 1]]}}
{{Implicatio... | Rule of Material Equivalence/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_2/Proof_1 | [
"Rule of Material Equivalence",
"Biconditional",
"Conditional",
"Conjunction"
] | [
"Rule of Material Equivalence",
"Definition:Valid Argument",
"Definition:Sequent",
"Definition:Propositional Logic",
"Rule of Material Equivalence/Formulation 1",
"Rule of Material Equivalence/Formulation 2"
] | [
"Rule of Material Equivalence/Formulation 1",
"Rule of Material Equivalence/Formulation 1"
] |
proofwiki-7325 | Rule of Material Equivalence | The '''Rule of Material Equivalence''' is a valid deduction sequent in propositional logic:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==== Formulation 1 ====
{{:Rule of Material Equivalence/Formulation 1}}
==== Formulation 2 ====
{... | 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|ccccccc|} \hline
(p & \iff & q) & \iff & (p & \implies & q) & \land & (q & \implies & p) \\
\hline
\F & \T & \F & \T & \F & \T & \F & \T ... | The '''[[Rule of Material Equivalence]]''' is a [[Definition:Valid Argument|valid]] deduction [[Definition:Sequent|sequent]] in [[Definition:Propositional Logic|propositional logic]]:
:If we can conclude that $p$ implies $q$ and if we can also conclude that $q$ implies $p$, then we may infer that $p$ {{iff}} $q$.
==... | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{... | Rule of Material Equivalence/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Material_Equivalence | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_2/Proof_by_Truth_Table | [
"Rule of Material Equivalence",
"Biconditional",
"Conditional",
"Conjunction"
] | [
"Rule of Material Equivalence",
"Definition:Valid Argument",
"Definition:Sequent",
"Definition:Propositional Logic",
"Rule of Material Equivalence/Formulation 1",
"Rule of Material Equivalence/Formulation 2"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-7326 | Union of Overlapping Order-Convex Sets in Toset is Order-Convex/Infinite Union | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\AA$ be a set of order-convex subsets of $S$.
For any $P, Q \in \AA$, let there be elements $C_0, \dotsc, C_n \in \AA$ such that:
:$C_0 = P$
:$C_n = Q$
:For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
Then $\bigcup \AA$ is order-convex in $S$. | Let $a, c \in \bigcup \AA$.
Let $b \in S$.
Let $a \prec b \prec c$.
Since $a, c \in \bigcup \AA$, there are $P, Q \in \AA$ such that $a \in P$ and $c \in Q$.
By the premise, there are elements $C_0, \dots, C_n \in \AA$ such that:
:$C_0 = P$
:$C_n = Q$
:For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
{{explain|det... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\AA$ be a [[Definition:Set of Sets|set]] of [[Definition:Order-Convex Set|order-convex]] [[Definition:Subset|subsets]] of $S$.
For any $P, Q \in \AA$, let there be elements $C_0, \dotsc, C_n \in \AA$ such that:
:$C_0 = P$
:$C... | Let $a, c \in \bigcup \AA$.
Let $b \in S$.
Let $a \prec b \prec c$.
Since $a, c \in \bigcup \AA$, there are $P, Q \in \AA$ such that $a \in P$ and $c \in Q$.
By the premise, there are elements $C_0, \dots, C_n \in \AA$ such that:
:$C_0 = P$
:$C_n = Q$
:For $k = 0, \dotsc, n - 1: C_k \cap C_{k + 1} \ne \O$
{{explai... | Union of Overlapping Order-Convex Sets in Toset is Order-Convex/Infinite Union | https://proofwiki.org/wiki/Union_of_Overlapping_Order-Convex_Sets_in_Toset_is_Order-Convex/Infinite_Union | https://proofwiki.org/wiki/Union_of_Overlapping_Order-Convex_Sets_in_Toset_is_Order-Convex/Infinite_Union | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Set of Sets",
"Definition:Order-Convex Set",
"Definition:Subset",
"Definition:Order-Convex Set"
] | [
"Union of Overlapping Order-Convex Sets in Toset is Order-Convex",
"Definition:Order-Convex Set",
"Definition:Order-Convex Set",
"Definition:Order-Convex Set",
"Category:Total Orderings"
] |
proofwiki-7327 | Rule of Material Equivalence/Formulation 2 | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} }$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} } }}
{{Assumption|1|p \iff q}}
{{SequentIntro|2|1|\paren {\paren {p \implies q} \land \paren {q \implies p} }|1|Rule of Material Equivalence: Formulation 1}}
{{Implication|3||\paren {p \iff q} \implies \paren {\paren ... | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} }$ | {{BeginTableau|\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} } }}
{{Assumption|1|p \iff q}}
{{SequentIntro|2|1|\paren {\paren {p \implies q} \land \paren {q \implies p} }|1|[[Rule of Material Equivalence/Formulation 1|Rule of Material Equivalence: Formulation 1]]}}
{{Implicatio... | Rule of Material Equivalence/Formulation 2/Proof 1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_2/Proof_1 | [
"Rule of Material Equivalence"
] | [] | [
"Rule of Material Equivalence/Formulation 1",
"Rule of Material Equivalence/Formulation 1"
] |
proofwiki-7328 | Rule of Material Equivalence/Formulation 2 | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} }$ | 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>$\begin{array}{|ccc|c|ccccccc|} \hline
(p & \iff & q) & \iff & (p & \implies & q) & \land & (q & \implies & p) \\
\hline
\F & \T & \F & \T & \F & \T & \F & \T ... | :$\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} }$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{... | Rule of Material Equivalence/Formulation 2/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_2/Proof_by_Truth_Table | [
"Rule of Material Equivalence"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-7329 | Continuous Involution is Homeomorphism | Let $\struct {S, \tau}$ be a topological space.
Let $f: S \to S$ be a continuous involution.
Then $f$ is a homeomorphism. | From Involution is Permutation, $f$ is a permutation and so a bijection.
Since $f$ is continuous, it suffices to verify that its inverse is also continuous.
Now recall $f$ is an involution, that is, $f^{-1} = f$.
Thus $f^{-1}$ is also continuous.
Hence $f$ is a homeomorphism.
{{qed}}
Category:Continuous Mappings
Catego... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $f: S \to S$ be a [[Definition:Continuous Mapping (Topology)|continuous]] [[Definition:Involution (Mapping)|involution]].
Then $f$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]]. | From [[Involution is Permutation]], $f$ is a [[Definition:Permutation|permutation]] and so a [[Definition:Bijection|bijection]].
Since $f$ is [[Definition:Continuous Mapping (Topology)|continuous]], it suffices to verify that its [[Definition:Inverse Mapping|inverse]] is also [[Definition:Continuous Mapping (Topology)... | Continuous Involution is Homeomorphism | https://proofwiki.org/wiki/Continuous_Involution_is_Homeomorphism | https://proofwiki.org/wiki/Continuous_Involution_is_Homeomorphism | [
"Continuous Mappings",
"Homeomorphisms (Topological Spaces)",
"Involutions"
] | [
"Definition:Topological Space",
"Definition:Continuous Mapping (Topology)",
"Definition:Involution (Mapping)",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Involution is Permutation",
"Definition:Permutation",
"Definition:Bijection",
"Definition:Continuous Mapping (Topology)",
"Definition:Inverse Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Involution (Mapping)",
"Definition:Continuous Mapping (Topology)",
"Definition:Homeomorphis... |
proofwiki-7330 | Involution is Permutation | Let $S$ be a set.
Let $f: S \to S$ be an involution.
Then $f$ is a permutation. | By definition, a permutation is a bijection from a set to itself.
Thus it is sufficient to show that $f$ is a bijection.
By definition of involution, for each $x \in S$:
:$\map f {\map f x} = x$
By Equality of Mappings:
:$f \circ f = I_S$
where $I_S$ is the identity mapping on $S$.
Thus $f$ is both a left inverse and a... | Let $S$ be a [[Definition:Set|set]].
Let $f: S \to S$ be an [[Definition:Involution (Mapping)|involution]].
Then $f$ is a [[Definition:Permutation|permutation]]. | By definition, a [[Definition:Permutation|permutation]] is a [[Definition:Bijection|bijection]] from a [[Definition:Set|set]] to itself.
Thus it is sufficient to show that $f$ is a [[Definition:Bijection|bijection]].
By definition of [[Definition:Involution (Mapping)|involution]], for each $x \in S$:
:$\map f {\map f... | Involution is Permutation | https://proofwiki.org/wiki/Involution_is_Permutation | https://proofwiki.org/wiki/Involution_is_Permutation | [
"Involutions",
"Permutations"
] | [
"Definition:Set",
"Definition:Involution (Mapping)",
"Definition:Permutation"
] | [
"Definition:Permutation",
"Definition:Bijection",
"Definition:Set",
"Definition:Bijection",
"Definition:Involution (Mapping)",
"Equality of Mappings",
"Definition:Identity Mapping",
"Definition:Left Inverse Mapping",
"Definition:Right Inverse Mapping",
"Bijection iff Left and Right Inverse",
"Ca... |
proofwiki-7331 | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication | : $\neg \left ({p \iff q}\right) \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|Rule of Material Equivalence}}
{{SequentIntro|3|1|\neg \paren {\paren {\neg p \lor q} \land ... | : $\neg \left ({p \iff q}\right) \vdash \left({\neg p \land q}\right) \lor \left({p \land \neg q}\right)$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Premise|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1
|[[Rule of Material Equivalence]]}}
{{SequentIntro|3|1|\neg \paren {\paren {\neg p \lor q} \l... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Forward_Implication/Proof | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Rule of Material Equivalence",
"Rule of Material Implication"
] |
proofwiki-7332 | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication | : $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right) \vdash \neg \left ({p \iff q}\right)$ | {{BeginTableau|\paren {\neg p \land q} \lor \paren {p \land \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise|1|\paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Commutation|2|1|\paren {p \land \neg q} \lor \paren {\neg p \land q}|1|Disjunction}}
{{Commutation|3|1|\paren {p \land \neg q} \lor \paren {q \land \... | : $\left({\neg p \land q}\right) \lor \left({p \land \neg q}\right) \vdash \neg \left ({p \iff q}\right)$ | {{BeginTableau|\paren {\neg p \land q} \lor \paren {p \land \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise|1|\paren {\neg p \land q} \lor \paren {p \land \neg q} }}
{{Commutation|2|1|\paren {p \land \neg q} \lor \paren {\neg p \land q}|1|Disjunction}}
{{Commutation|3|1|\paren {p \land \neg q} \lor \paren {q \land \... | Non-Equivalence as Disjunction of Conjunctions/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_1/Reverse_Implication/Proof | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Rule of Material Implication",
"Rule of Material Equivalence"
] |
proofwiki-7333 | Non-Equivalence as Disjunction of Conjunctions/Formulation 2 | :$\vdash \paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} }$ | {{BeginTableau|\vdash \paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} } }}
{{Assumption|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\paren {\neg p \land q} \lor \paren {p \land \neg q}|1|Non-Equivalence as Disjunction of Conjunctions: Formulation 1}}
{{Implication|3... | :$\vdash \paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} }$ | {{BeginTableau|\vdash \paren {\neg \paren {p \iff q} } \iff \paren {\paren {\neg p \land q} \lor \paren {p \land \neg q} } }}
{{Assumption|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|\paren {\neg p \land q} \lor \paren {p \land \neg q}|1|[[Non-Equivalence as Disjunction of Conjunctions/Formulation 1|Non-Equivalence ... | Non-Equivalence as Disjunction of Conjunctions/Formulation 2 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_2 | https://proofwiki.org/wiki/Non-Equivalence_as_Disjunction_of_Conjunctions/Formulation_2 | [
"Non-Equivalence as Disjunction of Conjunctions"
] | [] | [
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1",
"Non-Equivalence as Disjunction of Conjunctions/Formulation 1"
] |
proofwiki-7334 | Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication | :$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \iff \neg q} }}
{{Premise | 1|\neg \paren {p \iff q} }}
{{SequentIntro | 2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1|Rule of Material Equivalence}}
{{DeMorgan | 3|1|\neg \paren {p \implies q} \lor \neg \paren... | :$\neg \paren {p \iff q} \vdash \paren {p \iff \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \iff \neg q} }}
{{Premise | 1|\neg \paren {p \iff q} }}
{{SequentIntro | 2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1|[[Rule of Material Equivalence]]}}
{{DeMorgan | 3|1|\neg \paren {p \implies q} \lor \neg \p... | Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Forward_Implication/Proof | [
"Non-Equivalence as Equivalence with Negation"
] | [] | [
"Rule of Material Equivalence",
"Conjunction with Negative is Equivalent to Negation of Conditional",
"Rule of Transposition",
"Conjunction with Negative is Equivalent to Negation of Conditional"
] |
proofwiki-7335 | Non-Equivalence as Equivalence with Negation/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \iff \neg q} }}
{{Premise | 1|\neg \paren {p \iff q} }}
{{SequentIntro | 2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1|Rule of Material Equivalence}}
{{DeMorgan | 3|1|\neg \paren {p \implies q} \lor \neg \paren... | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | {{BeginTableau|\neg \paren {p \iff q} \vdash \paren {p \iff \neg q} }}
{{Premise | 1|\neg \paren {p \iff q} }}
{{SequentIntro | 2|1|\neg \paren {\paren {p \implies q} \land \paren {q \implies p} }|1|[[Rule of Material Equivalence]]}}
{{DeMorgan | 3|1|\neg \paren {p \implies q} \lor \neg \p... | Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Forward_Implication/Proof | [
"Non-Equivalence as Equivalence with Negation"
] | [] | [
"Rule of Material Equivalence",
"Conjunction with Negative is Equivalent to Negation of Conditional",
"Rule of Transposition",
"Conjunction with Negative is Equivalent to Negation of Conditional"
] |
proofwiki-7336 | Non-Equivalence as Equivalence with Negation/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | === Forward Implication: Proof ===
{{:Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication/Proof}}
=== Reverse Implication: Proof ===
{{:Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication/Proof}} | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | === [[Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication/Proof|Forward Implication: Proof]] ===
{{:Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication/Proof}}
=== [[Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication/Proof|Reverse Implica... | Non-Equivalence as Equivalence with Negation/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Proof_1 | [
"Non-Equivalence as Equivalence with Negation"
] | [] | [
"Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication/Proof",
"Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-7337 | Non-Equivalence as Equivalence with Negation/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccc|} \hline
\neg & (p & \iff & q) & (p & \iff & \neg & q) \\
\hline
\F & \F & \T & \F & \T & \F & \T & \F \\
\T & \F & \F & \T & \T & \T & \F & \T... | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|cccc||cccc|} \hline
\neg & (p... | Non-Equivalence as Equivalence with Negation/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Proof_by_Truth_Table | [
"Non-Equivalence as Equivalence with Negation"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7338 | Non-Equivalence as Equivalence with Negation/Formulation 1 | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | {{BeginTableau | \paren {p \iff \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise | 1 | p \iff \neg q}}
{{SequentIntro | 2 | 1 |\neg \paren {p \iff \neg \neg q} | 1 | Non-Equivalence as Equivalence with Negation: Forward Implication}}
{{TheoremIntro | 3 | \neg \neg q \iff q | Double Negation}}
{{SequentIntro | 4 ... | :$\neg \paren {p \iff q} \dashv \vdash \paren {p \iff \neg q}$ | {{BeginTableau | \paren {p \iff \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise | 1 | p \iff \neg q}}
{{SequentIntro | 2 | 1 |\neg \paren {p \iff \neg \neg q} | 1 | [[Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication|Non-Equivalence as Equivalence with Negation: Forward Implication]... | Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1 | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Reverse_Implication/Proof | [
"Non-Equivalence as Equivalence with Negation"
] | [] | [
"Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication",
"Double Negation",
"Biconditional is Transitive"
] |
proofwiki-7339 | Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication | :$\paren {p \iff \neg q} \vdash \neg \paren {p \iff q}$ | {{BeginTableau | \paren {p \iff \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise | 1 | p \iff \neg q}}
{{SequentIntro | 2 | 1 |\neg \paren {p \iff \neg \neg q} | 1 | Non-Equivalence as Equivalence with Negation: Forward Implication}}
{{TheoremIntro | 3 | \neg \neg q \iff q | Double Negation}}
{{SequentIntro | 4 ... | :$\paren {p \iff \neg q} \vdash \neg \paren {p \iff q}$ | {{BeginTableau | \paren {p \iff \neg q} \vdash \neg \paren {p \iff q} }}
{{Premise | 1 | p \iff \neg q}}
{{SequentIntro | 2 | 1 |\neg \paren {p \iff \neg \neg q} | 1 | [[Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication|Non-Equivalence as Equivalence with Negation: Forward Implication]... | Non-Equivalence as Equivalence with Negation/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_1/Reverse_Implication/Proof | [
"Non-Equivalence as Equivalence with Negation"
] | [] | [
"Non-Equivalence as Equivalence with Negation/Formulation 1/Forward Implication",
"Double Negation",
"Biconditional is Transitive"
] |
proofwiki-7340 | Sign of Function Matches Sign of Definite Integral | Let $f$ be a real function continuous on some closed interval $\closedint a b$, where $a < b$.
Then:
:If $\forall x \in \closedint a b: \map f x \ge 0$ then $\ds \int_a^b \map f x \rd x \ge 0$
:If $\forall x \in \closedint a b: \map f x > 0$ then $\ds \int_a^b \map f x \rd x > 0$
:If $\forall x \in \closedint a b: \map... | From Continuous Real Function is Darboux Integrable, the definite integrals under discussion are guaranteed to exist.
Consider the case where $\forall x \in \closedint a b: \map f x \ge 0$.
Define a constant mapping:
:$f_0: \closedint a b \to \R$:
:$\map {f_0} x = 0$
Then:
{{begin-eqn}}
{{eqn | l = \map {f_0} x
|... | Let $f$ be a [[Definition:Real Function|real function]] [[Definition:Continuous Real Function on Interval|continuous]] on some [[Definition:Closed Real Interval|closed interval]] $\closedint a b$, where $a < b$.
Then:
:If $\forall x \in \closedint a b: \map f x \ge 0$ then $\ds \int_a^b \map f x \rd x \ge 0$
:If $\f... | From [[Continuous Real Function is Darboux Integrable]], the [[Definition:Definite Integral|definite integrals]] under discussion are guaranteed to exist.
Consider the case where $\forall x \in \closedint a b: \map f x \ge 0$.
Define a [[Definition:Constant Mapping|constant mapping]]:
:$f_0: \closedint a b \to \R$:
... | Sign of Function Matches Sign of Definite Integral | https://proofwiki.org/wiki/Sign_of_Function_Matches_Sign_of_Definite_Integral | https://proofwiki.org/wiki/Sign_of_Function_Matches_Sign_of_Definite_Integral | [
"Integral Calculus"
] | [
"Definition:Real Function",
"Definition:Continuous Real Function/Interval",
"Definition:Real Interval/Closed"
] | [
"Continuous Real Function is Darboux Integrable",
"Definition:Definite Integral",
"Definition:Constant Mapping",
"Relative Sizes of Definite Integrals",
"Integral of Constant/Definite",
"Category:Integral Calculus"
] |
proofwiki-7341 | Non-Equivalence as Equivalence with Negation/Formulation 2 | :$\vdash \neg \paren {p \iff q} \iff \paren {p \iff \neg q}$ | {{BeginTableau|\vdash \neg \paren {p \iff q} \iff \paren {p \iff \neg q} }}
{{Assumption|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|p \iff \neg q|1|Non-Equivalence as Equivalence with Negation: Formulation 1}}
{{Implication|3||\paren {\neg \paren {p \iff q} } \implies \paren {p \iff \neg q}|1|2}}
{{Assumption|4|p \... | :$\vdash \neg \paren {p \iff q} \iff \paren {p \iff \neg q}$ | {{BeginTableau|\vdash \neg \paren {p \iff q} \iff \paren {p \iff \neg q} }}
{{Assumption|1|\neg \paren {p \iff q} }}
{{SequentIntro|2|1|p \iff \neg q|1|[[Non-Equivalence as Equivalence with Negation/Formulation 1|Non-Equivalence as Equivalence with Negation: Formulation 1]]}}
{{Implication|3||\paren {\neg \paren {p \if... | Non-Equivalence as Equivalence with Negation/Formulation 2 | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_2 | https://proofwiki.org/wiki/Non-Equivalence_as_Equivalence_with_Negation/Formulation_2 | [
"Non-Equivalence as Equivalence with Negation"
] | [] | [
"Non-Equivalence as Equivalence with Negation/Formulation 1",
"Non-Equivalence as Equivalence with Negation/Formulation 1"
] |
proofwiki-7342 | Transitive Relation whose Symmetric Closure is not Transitive | Let $S = \set {p, q}$, where $p$ and $q$ are distinct elements.
Let $\RR = \set {\tuple {p, q} }$.
Then $\RR$ is transitive but its symmetric closure is not. | $\RR$ is vacuously transitive because there are no elements $a, b, c \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR c$.
Let $\RR^\leftrightarrow$ be the symmetric closure of $\RR$.
Then $\RR^\leftrightarrow = \RR \cup \RR^{-1} = \set {\tuple {p, q}, \tuple {q, p} }$.
Then:
:$p \mathrel {\RR^\leftrightarrow} q$... | Let $S = \set {p, q}$, where $p$ and $q$ are distinct elements.
Let $\RR = \set {\tuple {p, q} }$.
Then $\RR$ is [[Definition:Transitive Relation|transitive]] but its [[Definition:Symmetric Closure|symmetric closure]] is not. | $\RR$ is vacuously [[Definition:Transitive Relation|transitive]] because there are no elements $a, b, c \in S$ such that $a \mathrel \RR b$ and $b \mathrel \RR c$.
Let $\RR^\leftrightarrow$ be the [[Definition:Symmetric Closure|symmetric closure]] of $\RR$.
Then $\RR^\leftrightarrow = \RR \cup \RR^{-1} = \set {\tuple... | Transitive Relation whose Symmetric Closure is not Transitive | https://proofwiki.org/wiki/Transitive_Relation_whose_Symmetric_Closure_is_not_Transitive | https://proofwiki.org/wiki/Transitive_Relation_whose_Symmetric_Closure_is_not_Transitive | [
"Transitive Relations",
"Symmetric Closures"
] | [
"Definition:Transitive Relation",
"Definition:Symmetric Closure"
] | [
"Definition:Transitive Relation",
"Definition:Symmetric Closure",
"Definition:Transitive Relation",
"Category:Transitive Relations",
"Category:Symmetric Closures"
] |
proofwiki-7343 | Symmetric Closure of Relation Compatible with Operation is Compatible | Let $\struct {S, \circ}$ be a magma.
Let $\RR$ be a relation compatible with $\circ$.
Let $\RR^\leftrightarrow$ denote the symmetric closure of $\RR$.
Then $\RR^\leftrightarrow$ is compatible with $\circ$. | By the definition of symmetric closure:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$.
Here $\RR^{-1}$ is the inverse of $\RR$.
By Inverse of Relation Compatible with Operation is Compatible, $\RR^{-1}$ is compatible with $\circ$.
Thus by Union of Relations Compatible with Operation is Compatible:
:$\RR^\leftrightarrow = ... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $\RR$ be a [[Definition:Endorelation|relation]] [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Let $\RR^\leftrightarrow$ denote the [[Definition:Symmetric Closure|symmetric closure]] of $\RR$.
Then $\RR^\leftrightarrow$ is [[D... | By the definition of [[Definition:Symmetric Closure|symmetric closure]]:
:$\RR^\leftrightarrow = \RR \cup \RR^{-1}$.
Here $\RR^{-1}$ is the [[Definition:Inverse Relation|inverse]] of $\RR$.
By [[Inverse of Relation Compatible with Operation is Compatible]], $\RR^{-1}$ is [[Definition:Relation Compatible with Operati... | Symmetric Closure of Relation Compatible with Operation is Compatible | https://proofwiki.org/wiki/Symmetric_Closure_of_Relation_Compatible_with_Operation_is_Compatible | https://proofwiki.org/wiki/Symmetric_Closure_of_Relation_Compatible_with_Operation_is_Compatible | [
"Compatible Relations",
"Symmetric Closures"
] | [
"Definition:Magma",
"Definition:Endorelation",
"Definition:Relation Compatible with Operation",
"Definition:Symmetric Closure",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Symmetric Closure",
"Definition:Inverse Relation",
"Inverse of Relation Compatible with Operation is Compatible",
"Definition:Relation Compatible with Operation",
"Union of Relations Compatible with Operation is Compatible",
"Definition:Relation Compatible with Operation",
"Category:Compatib... |
proofwiki-7344 | Singleton is Order-Convex Set | Let $\struct {S, \preceq}$ be an ordered set.
Let $x \in S$.
Then the singleton $\set x$ is order-convex. | Let:
:$a, c \in \set x$
:$b \in S$
:$a \preceq b \preceq c$
Then $a = c = x$.
Thus $x \preceq b \preceq x$.
Since $\preceq$ is a ordering, it is antisymmetric.
Thus $b = x$, so $b \in \set x$.
Since this holds for the only such triple, $\set x$ is order-convex.
{{qed}}
Category:Order-Convex Sets
Category:Singletons
cij... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $x \in S$.
Then the [[Definition:Singleton|singleton]] $\set x$ is [[Definition:Order-Convex Set|order-convex]]. | Let:
:$a, c \in \set x$
:$b \in S$
:$a \preceq b \preceq c$
Then $a = c = x$.
Thus $x \preceq b \preceq x$.
Since $\preceq$ is a [[Definition:Ordering|ordering]], it is [[Definition:Antisymmetric Relation|antisymmetric]].
Thus $b = x$, so $b \in \set x$.
Since this holds for the only such triple, $\set x$ is [[Def... | Singleton is Order-Convex Set | https://proofwiki.org/wiki/Singleton_is_Order-Convex_Set | https://proofwiki.org/wiki/Singleton_is_Order-Convex_Set | [
"Order-Convex Sets",
"Singletons"
] | [
"Definition:Ordered Set",
"Definition:Singleton",
"Definition:Order-Convex Set"
] | [
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Order-Convex Set",
"Category:Order-Convex Sets",
"Category:Singletons"
] |
proofwiki-7345 | Rule of Transposition/Variant 1/Formulation 2 | :$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$ | {{BeginTableau|\vdash \paren {p \implies \neg q} \implies \paren {q \implies \neg p} }}
{{Assumption|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{Implication|6||\paren {p \implies \neg q} \implies \paren {q \... | :$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$ | {{BeginTableau|\vdash \paren {p \implies \neg q} \implies \paren {q \implies \neg p} }}
{{Assumption|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{Implication|6||\paren {p \implies \neg q} \implies \paren {q \... | Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7346 | Rule of Transposition/Variant 1/Formulation 2 | :$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$ | === Proof of Forward Implication ===
{{:Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof}}
=== Proof of Reverse Implication ===
{{:Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof}}
{{BeginTableau|\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p} }}
{{Theore... | :$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$ | === [[Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof|Proof of Forward Implication]] ===
{{:Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof}}
=== [[Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof|Proof of Reverse Implication]] ===
{{:Rule of Trans... | Rule of Transposition/Variant 1/Formulation 2/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2/Proof | [
"Rule of Transposition"
] | [] | [
"Rule of Transposition/Variant 1/Formulation 2/Forward Implication/Proof",
"Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof",
"Rule of Transposition/Variant 1/Formulation 2/Forward Implication",
"Rule of Transposition/Variant 1/Formulation 2/Reverse Implication"
] |
proofwiki-7347 | Rule of Transposition/Variant 1/Formulation 2 | :$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$ | {{BeginTableau|\vdash \paren {q \implies \neg p} \implies \paren {p \implies \neg q} }}
{{Assumption|1|q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg q|1|3}}
{{Implication|5|1|p \implies \neg q|2|4}}
{{Implication|6||\paren {q \implies \neg p} \implies \paren {p \... | :$\vdash \paren {p \implies \neg q} \iff \paren {q \implies \neg p}$ | {{BeginTableau|\vdash \paren {q \implies \neg p} \implies \paren {p \implies \neg q} }}
{{Assumption|1|q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg q|1|3}}
{{Implication|5|1|p \implies \neg q|2|4}}
{{Implication|6||\paren {q \implies \neg p} \implies \paren {p \... | Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2/Reverse_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7348 | Rule of Transposition/Variant 1/Formulation 2/Reverse Implication | : $\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$ | {{BeginTableau|\vdash \paren {q \implies \neg p} \implies \paren {p \implies \neg q} }}
{{Assumption|1|q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg q|1|3}}
{{Implication|5|1|p \implies \neg q|2|4}}
{{Implication|6||\paren {q \implies \neg p} \implies \paren {p \... | : $\vdash \left({q \implies \neg p}\right) \implies \left({p \implies \neg q}\right)$ | {{BeginTableau|\vdash \paren {q \implies \neg p} \implies \paren {p \implies \neg q} }}
{{Assumption|1|q \implies \neg p}}
{{Assumption|2|p}}
{{DoubleNegIntro|3|2|\neg \neg p|2}}
{{ModusTollens|4|1, 2|\neg q|1|3}}
{{Implication|5|1|p \implies \neg q|2|4}}
{{Implication|6||\paren {q \implies \neg p} \implies \paren {p \... | Rule of Transposition/Variant 1/Formulation 2/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_2/Reverse_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7349 | Rule of Transposition/Variant 1/Formulation 1 | :$p \implies \neg q \dashv \vdash q \implies \neg p$ | {{BeginTableau|p \implies \neg q \vdash q \implies \neg p}}
{{Premise|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{EndTableau}}
{{Qed}} | :$p \implies \neg q \dashv \vdash q \implies \neg p$ | {{BeginTableau|p \implies \neg q \vdash q \implies \neg p}}
{{Premise|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{EndTableau}}
{{Qed}} | Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7350 | Rule of Transposition/Variant 1/Formulation 1 | :$p \implies \neg q \dashv \vdash q \implies \neg p$ | === Proof of Forward Implication ===
{{:Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof}}
=== Proof of Reverse Implication ===
{{:Rule of Transposition/Variant 1/Formulation 1/Reverse Implication/Proof}} | :$p \implies \neg q \dashv \vdash q \implies \neg p$ | === [[Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof|Proof of Forward Implication]] ===
{{:Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof}}
=== [[Rule of Transposition/Variant 1/Formulation 1/Reverse Implication/Proof|Proof of Reverse Implication]] ===
{{:Rule of Trans... | Rule of Transposition/Variant 1/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1/Proof_1 | [
"Rule of Transposition"
] | [] | [
"Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof",
"Rule of Transposition/Variant 1/Formulation 1/Reverse Implication/Proof"
] |
proofwiki-7351 | Rule of Transposition/Variant 1/Formulation 1 | :$p \implies \neg q \dashv \vdash q \implies \neg p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
:<nowiki>$\begin{array}{|cccc||cccc|} \hline
p & \implies & \neg & q & q & \implies & \neg & p \\
\hline
\F & \T & \T & \F & \F & \T & \T & \F \\
\F & ... | :$p \implies \neg q \dashv \vdash q \implies \neg p$ | We apply the [[Method of Truth Tables]] to the proposition.
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{|cc... | Rule of Transposition/Variant 1/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7352 | Rule of Transposition/Variant 1/Formulation 1/Forward Implication | :$p \implies \neg q \vdash q \implies \neg p$ | {{BeginTableau|p \implies \neg q \vdash q \implies \neg p}}
{{Premise|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{EndTableau}}
{{Qed}} | :$p \implies \neg q \vdash q \implies \neg p$ | {{BeginTableau|p \implies \neg q \vdash q \implies \neg p}}
{{Premise|1|p \implies \neg q}}
{{Assumption|2|q}}
{{DoubleNegIntro|3|2|\neg \neg q|2}}
{{ModusTollens|4|1, 2|\neg p|1|3}}
{{Implication|5|1|q \implies \neg p|2|4}}
{{EndTableau}}
{{Qed}} | Rule of Transposition/Variant 1/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_1/Formulation_1/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7353 | Rule of Transposition/Variant 2/Formulation 1 | : $\neg p \implies q \dashv \vdash \neg q \implies p$ | {{BeginTableau|\neg p \implies q \vdash \neg q \implies p}}
{{Premise|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | : $\neg p \implies q \dashv \vdash \neg q \implies p$ | {{BeginTableau|\neg p \implies q \vdash \neg q \implies p}}
{{Premise|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | Rule of Transposition/Variant 2/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7354 | Rule of Transposition/Variant 2/Formulation 1 | : $\neg p \implies q \dashv \vdash \neg q \implies p$ | We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccc||cccc|} \hline
\neg & p & \implies & q & \neg & q & \implies & p \\
\hline
\T & \F & \T & \F & \T & \F & \T & \F \\
\T & \F & \T &... | : $\neg p \implies q \dashv \vdash \neg q \implies p$ | 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 connectives]] match for all [[Definition:Boolean Interpretation|boolean interpretations]].
$\begin{array}{|cccc||cccc... | Rule of Transposition/Variant 2/Formulation 1/Proof by Truth Table | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Proof_by_Truth_Table | [
"Rule of Transposition"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7355 | Rule of Transposition/Variant 2/Formulation 1 | : $\neg p \implies q \dashv \vdash \neg q \implies p$ | {{BeginTableau|\neg q \implies p \vdash \neg p \implies q}}
{{Premise|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | : $\neg p \implies q \dashv \vdash \neg q \implies p$ | {{BeginTableau|\neg q \implies p \vdash \neg p \implies q}}
{{Premise|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | Rule of Transposition/Variant 2/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Reverse_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7356 | Rule of Transposition/Variant 2/Formulation 1/Forward Implication | :$\neg p \implies q \vdash \neg q \implies p$ | {{BeginTableau|\neg p \implies q \vdash \neg q \implies p}}
{{Premise|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | :$\neg p \implies q \vdash \neg q \implies p$ | {{BeginTableau|\neg p \implies q \vdash \neg q \implies p}}
{{Premise|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | Rule of Transposition/Variant 2/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7357 | Rule of Transposition/Variant 2/Formulation 1/Forward Implication/Proof | : $\neg p \implies q \vdash \neg q \implies p$ | {{BeginTableau|\neg p \implies q \vdash \neg q \implies p}}
{{Premise|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | : $\neg p \implies q \vdash \neg q \implies p$ | {{BeginTableau|\neg p \implies q \vdash \neg q \implies p}}
{{Premise|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | Rule of Transposition/Variant 2/Formulation 1/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Forward_Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7358 | Rule of Transposition/Variant 2/Formulation 1/Reverse Implication | : $q \implies \neg p \vdash p \implies \neg q$ | {{BeginTableau|\neg q \implies p \vdash \neg p \implies q}}
{{Premise|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | : $q \implies \neg p \vdash p \implies \neg q$ | {{BeginTableau|\neg q \implies p \vdash \neg p \implies q}}
{{Premise|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | Rule of Transposition/Variant 2/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Reverse_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7359 | Rule of Transposition/Variant 2/Formulation 1/Reverse Implication/Proof | : $\neg q \implies p \vdash \neg p \implies q$ | {{BeginTableau|\neg q \implies p \vdash \neg p \implies q}}
{{Premise|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | : $\neg q \implies p \vdash \neg p \implies q$ | {{BeginTableau|\neg q \implies p \vdash \neg p \implies q}}
{{Premise|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{EndTableau|qed}}
{{LEM|Double Negation Elimination|3}} | Rule of Transposition/Variant 2/Formulation 1/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Reverse_Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_1/Reverse_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7360 | Rule of Transposition/Variant 2/Formulation 2/Forward Implication | :$\vdash \paren {\neg p \implies q} \implies \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg p \implies q} \implies \paren {\neg q \implies p} }}
{{Assumption|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{Implication|6||\paren {\neg p \implies q} \implies \p... | :$\vdash \paren {\neg p \implies q} \implies \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg p \implies q} \implies \paren {\neg q \implies p} }}
{{Assumption|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{Implication|6||\paren {\neg p \implies q} \implies \p... | Rule of Transposition/Variant 2/Formulation 2/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2/Forward_Implication | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7361 | Rule of Transposition/Variant 2/Formulation 2/Reverse Implication | :$\vdash \paren {\neg q \implies p} \implies \paren {\neg p \implies q}$ | {{BeginTableau|\vdash \paren {\neg q \implies p} \implies \paren {\neg p \implies q} }}
{{Assumption|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{Implication|6||\paren {\neg q \implies p} \implies \p... | :$\vdash \paren {\neg q \implies p} \implies \paren {\neg p \implies q}$ | {{BeginTableau|\vdash \paren {\neg q \implies p} \implies \paren {\neg p \implies q} }}
{{Assumption|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{Implication|6||\paren {\neg q \implies p} \implies \p... | Rule of Transposition/Variant 2/Formulation 2/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2/Reverse_Implication | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2/Reverse_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7362 | Rule of Transposition/Variant 2/Formulation 2 | :$\vdash \paren {\neg p \implies q}\ \iff \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg p \implies q} \implies \paren {\neg q \implies p} }}
{{Assumption|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{Implication|6||\paren {\neg p \implies q} \implies \p... | :$\vdash \paren {\neg p \implies q}\ \iff \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg p \implies q} \implies \paren {\neg q \implies p} }}
{{Assumption|1|\neg p \implies q}}
{{Assumption|2|\neg q}}
{{ModusTollens|3|1, 2|\neg \neg p|1|2}}
{{DoubleNegElimination|4|1, 2|p|3}}
{{Implication|5|1|\neg q \implies p|2|4}}
{{Implication|6||\paren {\neg p \implies q} \implies \p... | Rule of Transposition/Variant 2/Formulation 2/Forward Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2/Forward_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7363 | Rule of Transposition/Variant 2/Formulation 2 | :$\vdash \paren {\neg p \implies q}\ \iff \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg p \implies q} \iff \paren {\neg q \implies p} }}
{{TheoremIntro|1|\paren {\neg p \implies q} \implies \paren {\neg q \implies p}|Rule of Transposition: Forward Implication}}
{{TheoremIntro|2|\paren {\neg q \implies p} \implies \paren {\neg p \implies q}|Rule of Transposition: Reverse ... | :$\vdash \paren {\neg p \implies q}\ \iff \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg p \implies q} \iff \paren {\neg q \implies p} }}
{{TheoremIntro|1|\paren {\neg p \implies q} \implies \paren {\neg q \implies p}|[[Rule of Transposition/Variant 2/Formulation 2/Forward Implication|Rule of Transposition: Forward Implication]]}}
{{TheoremIntro|2|\paren {\neg q \implies ... | Rule of Transposition/Variant 2/Formulation 2/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2/Proof | [
"Rule of Transposition"
] | [] | [
"Rule of Transposition/Variant 2/Formulation 2/Forward Implication",
"Rule of Transposition/Variant 2/Formulation 2/Reverse Implication"
] |
proofwiki-7364 | Rule of Transposition/Variant 2/Formulation 2 | :$\vdash \paren {\neg p \implies q}\ \iff \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg q \implies p} \implies \paren {\neg p \implies q} }}
{{Assumption|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{Implication|6||\paren {\neg q \implies p} \implies \p... | :$\vdash \paren {\neg p \implies q}\ \iff \paren {\neg q \implies p}$ | {{BeginTableau|\vdash \paren {\neg q \implies p} \implies \paren {\neg p \implies q} }}
{{Assumption|1|\neg q \implies p}}
{{Assumption|2|\neg p}}
{{ModusTollens|3|1, 2|\neg \neg q|1|2}}
{{DoubleNegElimination|4|1, 2|q|3}}
{{Implication|5|1|\neg p \implies q|2|4}}
{{Implication|6||\paren {\neg q \implies p} \implies \p... | Rule of Transposition/Variant 2/Formulation 2/Reverse Implication/Proof | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2 | https://proofwiki.org/wiki/Rule_of_Transposition/Variant_2/Formulation_2/Reverse_Implication/Proof | [
"Rule of Transposition"
] | [] | [] |
proofwiki-7365 | Boundary of Polygon is Topological Boundary | Let $P$ be a polygon embedded in $\R^2$.
Denote the boundary of $P$ as $\partial P$.
Let $\Int P$ and $\Ext P$ denote the interior and exterior of $\partial P$, where $\partial P$ is considered as a Jordan curve.
Then the topological boundary of $\Int P$ is equal to $\partial P$, and the topological boundary of $\Ext P... | Denote the topological boundary of $\Int P$ as $\partial \Int P$, and denote the topological boundary of $\Ext P$ as $\partial \Ext P$. | Let $P$ be a [[Definition:Polygon|polygon]] embedded in $\R^2$.
Denote the [[Definition:Boundary (Geometry)|boundary]] of $P$ as $\partial P$.
Let $\Int P$ and $\Ext P$ denote the [[Definition:Interior of Jordan Curve|interior]] and [[Definition:Exterior of Jordan Curve|exterior]] of $\partial P$, where $\partial P$ ... | Denote the [[Definition:Boundary (Topology)|topological boundary]] of $\Int P$ as $\partial \Int P$, and denote the [[Definition:Boundary (Topology)|topological boundary]] of $\Ext P$ as $\partial \Ext P$. | Boundary of Polygon is Topological Boundary | https://proofwiki.org/wiki/Boundary_of_Polygon_is_Topological_Boundary | https://proofwiki.org/wiki/Boundary_of_Polygon_is_Topological_Boundary | [
"Plane Geometry",
"Topology"
] | [
"Definition:Polygon",
"Definition:Boundary (Geometry)",
"Definition:Jordan Curve/Interior",
"Definition:Jordan Curve/Exterior",
"Definition:Jordan Curve",
"Definition:Boundary (Topology)",
"Definition:Boundary (Topology)"
] | [
"Definition:Boundary (Topology)",
"Definition:Boundary (Topology)"
] |
proofwiki-7366 | Convex Component of Open Set in GO-Space is Open | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space.
Let $A$ be a $\tau$-open subset of $S$.
Let $C$ be a convex component of $A$ in $S$
Then $C$ is open relative to $\tau$. | Note: the term '''convex''' will be used here to refer to a set that is order-convex in $S$.
Since $\struct {S, \preceq, \tau}$ is a generalized ordered space, $\tau$ has a basis $\BB$ consisting of open convex sets.
Let $x \in C$.
Then $x \in A$, and $A$ is open in $S$.
Thus by the definition of a basis, there is a $U... | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space|generalized ordered space]].
Let $A$ be a $\tau$-open subset of $S$.
Let $C$ be a convex component of $A$ in $S$
Then $C$ is open relative to $\tau$. | Note: the term '''convex''' will be used here to refer to a set that is [[Definition:Order-Convex Set|order-convex]] in $S$.
Since $\struct {S, \preceq, \tau}$ is a [[Definition:Generalized Ordered Space|generalized ordered space]], $\tau$ has a [[Definition:Basis (Topology)|basis]] $\BB$ consisting of open convex set... | Convex Component of Open Set in GO-Space is Open | https://proofwiki.org/wiki/Convex_Component_of_Open_Set_in_GO-Space_is_Open | https://proofwiki.org/wiki/Convex_Component_of_Open_Set_in_GO-Space_is_Open | [
"Generalized Ordered Spaces",
"Total Orderings"
] | [
"Definition:Generalized Ordered Space"
] | [
"Definition:Order-Convex Set",
"Definition:Generalized Ordered Space",
"Definition:Basis (Topology)",
"Category:Generalized Ordered Spaces",
"Category:Total Orderings"
] |
proofwiki-7367 | Jordan Curve and Jordan Arc form Two Jordan Curves | Let $\gamma: \closedint a b \to \R^2$ be a Jordan curve, where $\closedint a b$ is the closed real interval between $a, b \in \R$ with $a < b$.
Let the interior of $\gamma$ be denoted $\Int \gamma$.
Let the image of $\gamma$ be denoted $\Img \gamma$.
Let $\sigma: \closedint c d \to \R^2$ be a Jordan arc such that:
:$\m... | === $\gamma_1$ and $\gamma_2$ are Jordan curves ===
Let $\closedint {a_1}{b_1}$ denote the domain of $\gamma_1$, and $\closedint {a_2}{b_2}$ denote the domain of $\gamma_2$.
As:
:$\Int \gamma$ and $\Img \gamma$ are disjoint by the Jordan Curve Theorem
and:
:$\map \sigma {\openint c d} \subseteq \Int \gamma$
it follows ... | Let $\gamma: \closedint a b \to \R^2$ be a [[Definition:Jordan Curve|Jordan curve]], where $\closedint a b$ is the [[Definition:Closed Real Interval|closed real interval]] between $a, b \in \R$ with $a < b$.
Let the [[Definition:Interior of Jordan Curve|interior]] of $\gamma$ be denoted $\Int \gamma$.
Let the [[Defin... | === $\gamma_1$ and $\gamma_2$ are [[Definition:Jordan Curve|Jordan curves]] ===
Let $\closedint {a_1}{b_1}$ denote the [[Definition:Domain of Mapping|domain]] of $\gamma_1$, and $\closedint {a_2}{b_2}$ denote the [[Definition:Domain of Mapping|domain]] of $\gamma_2$.
As:
:$\Int \gamma$ and $\Img \gamma$ are [[Definit... | Jordan Curve and Jordan Arc form Two Jordan Curves | https://proofwiki.org/wiki/Jordan_Curve_and_Jordan_Arc_form_Two_Jordan_Curves | https://proofwiki.org/wiki/Jordan_Curve_and_Jordan_Arc_form_Two_Jordan_Curves | [
"Jordan Curves",
"Jordan Arcs"
] | [
"Definition:Jordan Curve",
"Definition:Real Interval/Closed",
"Definition:Jordan Curve/Interior",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Jordan Arc",
"Definition:Concatenation (Topology)",
"Definition:Restriction/Mapping",
"Definition:Jordan Curve"
] | [
"Definition:Jordan Curve",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Disjoint Sets",
"Jordan Curve Theorem",
"Definition:Jordan Curve",
"Definition:Set Intersection",
"Definition:Path (Topology)/Initial Point",
"Definition:Path (Topology)/Final P... |
proofwiki-7368 | Restriction of Total Ordering is Total Ordering | Let $\struct {S, \preceq}$ be a total ordering.
Let $T \subseteq S$.
Let $\preceq \restriction_T$ be the restriction of $\preceq$ to $T$.
Then $\preceq \restriction_T$ is a total ordering of $T$. | By Restriction of Ordering is Ordering, $\preceq \restriction_T$ is an ordering.
Let $x, y \in T$.
As $T \subseteq S$ it follows by definition of subset that:
:$x, y \in S$
As $\preceq$ is a total ordering:
:$\tuple {x, y} \in {\preceq}$
or:
:$\tuple {y, x} \in {\preceq}$
Suppose $\tuple {x, y} \in {\preceq}$.
As $x, y... | Let $\struct {S, \preceq}$ be a [[Definition:Total Ordering|total ordering]].
Let $T \subseteq S$.
Let $\preceq \restriction_T$ be the [[Definition:Restriction of Ordering|restriction]] of $\preceq$ to $T$.
Then $\preceq \restriction_T$ is a [[Definition:Total Ordering|total ordering]] of $T$. | By [[Restriction of Ordering is Ordering]], $\preceq \restriction_T$ is an [[Definition:Ordering|ordering]].
Let $x, y \in T$.
As $T \subseteq S$ it follows by definition of [[Definition:Subset|subset]] that:
:$x, y \in S$
As $\preceq$ is a [[Definition:Total Ordering|total ordering]]:
:$\tuple {x, y} \in {\preceq}$... | Restriction of Total Ordering is Total Ordering | https://proofwiki.org/wiki/Restriction_of_Total_Ordering_is_Total_Ordering | https://proofwiki.org/wiki/Restriction_of_Total_Ordering_is_Total_Ordering | [
"Total Orderings"
] | [
"Definition:Total Ordering",
"Definition:Restriction of Ordering",
"Definition:Total Ordering"
] | [
"Restriction of Ordering is Ordering",
"Definition:Ordering",
"Definition:Subset",
"Definition:Total Ordering",
"Definition:Cartesian Product",
"Definition:Restriction of Ordering",
"Definition:Total Ordering"
] |
proofwiki-7369 | Ordering Cycle implies Equality/General Case | Let $\struct {S,\preceq}$ be an ordered set.
Let $x_0, x_1, \dots, x_n \in S$.
Suppose that for $k = 0, 1, \dots, n - 1: x_k \preceq x_{k + 1}$.
Suppose also that $x_n \preceq x_0$.
Then $x_0 = x_1 = \dots = x_n$. | Since $\preceq$ is an ordering it is transitive and antisymmetric.
By Transitive Chaining, it follows from the first premise that for all $k$ with $0 \le k \le n$:
:$x_0 \preceq x_k$
and also:
:$x_k \preceq x_n$
The other premise states that $x_n \preceq x_0$.
By transitivity of $\preceq$, this combines with the above ... | Let $\struct {S,\preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $x_0, x_1, \dots, x_n \in S$.
Suppose that for $k = 0, 1, \dots, n - 1: x_k \preceq x_{k + 1}$.
Suppose also that $x_n \preceq x_0$.
Then $x_0 = x_1 = \dots = x_n$. | Since $\preceq$ is an [[Definition:Ordering|ordering]] it is [[Definition:Transitive Relation|transitive]] and [[Definition:Antisymmetric Relation|antisymmetric]].
By [[Transitive Chaining]], it follows from the first premise that for all $k$ with $0 \le k \le n$:
:$x_0 \preceq x_k$
and also:
:$x_k \preceq x_n$
Th... | Ordering Cycle implies Equality/General Case | https://proofwiki.org/wiki/Ordering_Cycle_implies_Equality/General_Case | https://proofwiki.org/wiki/Ordering_Cycle_implies_Equality/General_Case | [
"Order Theory"
] | [
"Definition:Ordered Set"
] | [
"Definition:Ordering",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Transitive Chaining",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation"
] |
proofwiki-7370 | Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a finite, non-empty subset of $S$.
Then $T$ has a maximal element and a minimal element. | We will show that each finite subset of $S$ has a minimal element.
The existence of a maximal element then follows from duality.
Proof by induction:
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:Every set with $n$ elements has a minimal element. | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$ be a [[Definition:Finite Set|finite]], [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
Then $T$ has a [[Definition:Maximal Element|maximal element]] and a [[Definition:Minimal Element|minimal elem... | We will show that each finite subset of $S$ has a [[Definition:Minimal Element|minimal element]].
The existence of a [[Definition:Maximal Element|maximal element]] then follows from duality.
Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the [[Definition:Pr... | Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements | https://proofwiki.org/wiki/Finite_Non-Empty_Subset_of_Ordered_Set_has_Maximal_and_Minimal_Elements | https://proofwiki.org/wiki/Finite_Non-Empty_Subset_of_Ordered_Set_has_Maximal_and_Minimal_Elements | [
"Minimal Elements",
"Maximal Elements"
] | [
"Definition:Ordered Set",
"Definition:Finite Set",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Maximal/Element",
"Definition:Minimal/Element"
] | [
"Definition:Minimal/Element",
"Definition:Maximal/Element",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Minimal/Ele... |
proofwiki-7371 | Star Convex Set is Path-Connected | Let $A$ be a star convex subset of a topological vector space $V$ over $\R$ or $\C$.
Then $A$ is path-connected. | Let $x_1, x_2 \in A$.
Let $a \in A$ be a star center of $A$.
By definition of star convex set:
:$\forall t \in \closedint 0 1: x_1 + \paren {1 - t} a, t x_2 + \paren {1 - t} a \in A$
Define two paths $\gamma_1, \gamma_2: t \in \closedint 0 1 \to A$ by:
:$\map {\gamma_1} t = t x_1 + \paren {1 - t} a$
:$\map {\gamma_2} ... | Let $A$ be a [[Definition:Star Convex Set|star convex]] [[Definition:Subset|subset]] of a [[Definition:Topological Vector Space|topological vector space]] $V$ over $\R$ or $\C$.
Then $A$ is [[Definition:Path-Connected Set|path-connected]]. | Let $x_1, x_2 \in A$.
Let $a \in A$ be a [[Definition:Star Center|star center]] of $A$.
By definition of [[Definition:Star Convex Set|star convex set]]:
:$\forall t \in \closedint 0 1: x_1 + \paren {1 - t} a, t x_2 + \paren {1 - t} a \in A$
Define two [[Definition:Path (Topology)|paths]] $\gamma_1, \gamma_2: t \in... | Star Convex Set is Path-Connected | https://proofwiki.org/wiki/Star_Convex_Set_is_Path-Connected | https://proofwiki.org/wiki/Star_Convex_Set_is_Path-Connected | [
"Vector Spaces",
"Path-Connected Sets"
] | [
"Definition:Star Convex Set",
"Definition:Subset",
"Definition:Topological Vector Space",
"Definition:Path-Connected/Set"
] | [
"Definition:Star Convex Set/Star Center",
"Definition:Star Convex Set",
"Definition:Path (Topology)",
"Definition:Topological Vector Space",
"Definition:Continuous Mapping (Topology)/Set",
"Definition:Concatenation (Topology)",
"Definition:Path (Topology)",
"Definition:Path-Connected/Set"
] |
proofwiki-7372 | Minimal Element of Chain is Smallest Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a chain in $S$.
Let $m$ be a minimal element of $C$.
Then $m$ is the smallest element of $C$. | Let $x \in C$.
Since $m$ is minimal in $C$, $x \nprec m$.
Since $C$ is a chain, $x = m$ or $m \prec x$.
Thus for each $x \in C$, $m \preceq x$.
Therefore $m$ is the smallest element of $C$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $C$ be a [[Definition:Chain (Order Theory)|chain]] in $S$.
Let $m$ be a [[Definition:Minimal Element|minimal element]] of $C$.
Then $m$ is the [[Definition:Smallest Element|smallest element]] of $C$. | Let $x \in C$.
Since $m$ is [[Definition:Minimal Element|minimal]] in $C$, $x \nprec m$.
Since $C$ is a [[Definition:Chain (Order Theory)|chain]], $x = m$ or $m \prec x$.
Thus for each $x \in C$, $m \preceq x$.
Therefore $m$ is the [[Definition:Smallest Element|smallest element]] of $C$.
{{qed}} | Minimal Element of Chain is Smallest Element | https://proofwiki.org/wiki/Minimal_Element_of_Chain_is_Smallest_Element | https://proofwiki.org/wiki/Minimal_Element_of_Chain_is_Smallest_Element | [
"Minimal Elements",
"Smallest Elements"
] | [
"Definition:Ordered Set",
"Definition:Chain (Order Theory)",
"Definition:Minimal/Element",
"Definition:Smallest Element"
] | [
"Definition:Minimal/Element",
"Definition:Chain (Order Theory)",
"Definition:Smallest Element"
] |
proofwiki-7373 | Maximal Element of Chain is Greatest Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a chain in $S$.
Let $m$ be a maximal element of $C$.
Then $m$ is the greatest element of $C$. | Let $x \in C$.
Since $m$ is maximal in $C$, $m \nprec x$.
Since $C$ is a chain, $x = m$ or $x \prec m$.
Thus for each $x \in C$, $x \preceq m$.
Therefore $m$ is the greatest element of $C$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $C$ be a [[Definition:Chain (Order Theory)|chain]] in $S$.
Let $m$ be a [[Definition:Maximal Element|maximal element]] of $C$.
Then $m$ is the [[Definition:Greatest Element|greatest element]] of $C$. | Let $x \in C$.
Since $m$ is [[Definition:Maximal Element|maximal]] in $C$, $m \nprec x$.
Since $C$ is a [[Definition:Chain (Order Theory)|chain]], $x = m$ or $x \prec m$.
Thus for each $x \in C$, $x \preceq m$.
Therefore $m$ is the [[Definition:Greatest Element|greatest element]] of $C$.
{{qed}} | Maximal Element of Chain is Greatest Element | https://proofwiki.org/wiki/Maximal_Element_of_Chain_is_Greatest_Element | https://proofwiki.org/wiki/Maximal_Element_of_Chain_is_Greatest_Element | [
"Maximal Elements",
"Greatest Elements",
"Chains (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Chain (Order Theory)",
"Definition:Maximal/Element",
"Definition:Greatest Element"
] | [
"Definition:Maximal/Element",
"Definition:Chain (Order Theory)",
"Definition:Greatest Element"
] |
proofwiki-7374 | Modus Ponendo Ponens/Variant 3 | :$\vdash \paren {\paren {p \implies q} \land p} \implies q$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \land p} \implies q}}
{{Assumption|1|\paren {p \implies q} \land p}}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|p|1|2}}
{{ModusPonens|4|1|q|2|3}}
{{Implication|5||\paren {\paren {p \implies q} \land p} \implies q|1|4}}
{{EndTableau}}
{{Qed}} | :$\vdash \paren {\paren {p \implies q} \land p} \implies q$ | {{BeginTableau|\vdash \paren {\paren {p \implies q} \land p} \implies q}}
{{Assumption|1|\paren {p \implies q} \land p}}
{{Simplification|2|1|p \implies q|1|1}}
{{Simplification|3|1|p|1|2}}
{{ModusPonens|4|1|q|2|3}}
{{Implication|5||\paren {\paren {p \implies q} \land p} \implies q|1|4}}
{{EndTableau}}
{{Qed}} | Modus Ponendo Ponens/Variant 3/Proof 1 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_3 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_3/Proof_1 | [
"Modus Ponendo Ponens"
] | [] | [] |
proofwiki-7375 | Modus Ponendo Ponens/Variant 3 | :$\vdash \paren {\paren {p \implies q} \land p} \implies q$ | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective are true for all boolean interpretations.
:<nowiki>$\begin{array}{|ccccc|c|c|} \hline
((p & \implies & q) & \land & p) & \implies & q \\
\hline
\F & \T & \F & \F & \F & \T & \F \\
\F & \T & \T & \F & \F & \T & ... | :$\vdash \paren {\paren {p \implies q} \land p} \implies q$ | We apply the [[Method of Truth Tables]].
As can be seen by inspection, the [[Definition:Truth Value|truth values]] under the [[Definition:Main Connective (Propositional Logic)|main connective]] are [[Definition:True|true]] for all [[Definition:Boolean Interpretation|boolean interpretations]].
:<nowiki>$\begin{array}{... | Modus Ponendo Ponens/Variant 3/Proof by Truth Table | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_3 | https://proofwiki.org/wiki/Modus_Ponendo_Ponens/Variant_3/Proof_by_Truth_Table | [
"Modus Ponendo Ponens"
] | [] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:True",
"Definition:Boolean Interpretation"
] |
proofwiki-7376 | Rule of Material Equivalence/Formulation 1/Proof 1 | :$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$ | {{BeginTableau|p \iff q \vdash \paren {p \implies q} \land \paren {q \implies p} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{Conjunction|4|1|\paren {p \implies q} \land \paren {q \implies p}|2|3}}
{{EndTableau}}
{{BeginTableau|\paren {p... | :$p \iff q \dashv \vdash \paren {p \implies q} \land \paren {q \implies p}$ | {{BeginTableau|p \iff q \vdash \paren {p \implies q} \land \paren {q \implies p} }}
{{Premise|1|p \iff q}}
{{BiconditionalElimination|2|1|p \implies q|1|1}}
{{BiconditionalElimination|3|1|q \implies p|1|2}}
{{Conjunction|4|1|\paren {p \implies q} \land \paren {q \implies p}|2|3}}
{{EndTableau}}
{{BeginTableau|\paren ... | Rule of Material Equivalence/Formulation 1/Proof 1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1/Proof_1 | https://proofwiki.org/wiki/Rule_of_Material_Equivalence/Formulation_1/Proof_1 | [
"Rule of Material Equivalence"
] | [] | [] |
proofwiki-7377 | Order-Convex Set Characterization | Let $\struct {S, \preceq, \tau}$ be an ordered set.
Let $C \subseteq S$.
{{TFAE}}
{{begin-axiom}}
{{axiom | n = 1
| t = $C$ is order-convex.
}}
{{axiom | n = 2
| t = $C$ is the intersection of an upper section with a lower section.
}}
{{axiom | n = 3
| t = $C$ is the intersection of its upper cl... | === $(2)$ implies $(1)$ ===
Follows from Upper Section is Order-Convex, Lower Section is Order-Convex, and Intersection of Order-Convex Sets is Order-Convex.
{{qed|lemma}} | Let $\struct {S, \preceq, \tau}$ be an [[Definition:Ordered Set|ordered set]].
Let $C \subseteq S$.
{{TFAE}}
{{begin-axiom}}
{{axiom | n = 1
| t = $C$ is [[Definition:Order-Convex Set|order-convex]].
}}
{{axiom | n = 2
| t = $C$ is the intersection of an [[Definition:Upper Section|upper section]] wit... | === $(2)$ implies $(1)$ ===
Follows from [[Upper Section is Order-Convex]], [[Lower Section is Order-Convex]], and [[Intersection of Order-Convex Sets is Order-Convex]].
{{qed|lemma}} | Order-Convex Set Characterization | https://proofwiki.org/wiki/Order-Convex_Set_Characterization | https://proofwiki.org/wiki/Order-Convex_Set_Characterization | [
"Order-Convex Sets"
] | [
"Definition:Ordered Set",
"Definition:Order-Convex Set",
"Definition:Upper Section",
"Definition:Lower Section"
] | [
"Upper Section is Order-Convex",
"Lower Section is Order-Convex",
"Intersection of Order-Convex Sets is Order-Convex"
] |
proofwiki-7378 | Upper Closure is Upper Section | Let $\struct {S, \preceq}$ be an ordered set.
Let $T$ be a subset of $S$.
Let $U$ be the upper closure of $T$.
Then $U$ is an upper section. | Let $a \in U$.
Let $b \in S$ with $a \preceq b$.
By the definition of upper closure, there is a $t \in T$ such that $t \preceq a$.
By transitivity, $t \preceq b$.
Thus, again by the definition of upper closure:
:$b \in U$
Since this holds for all such $a$ and $b$, $U$ is an upper section.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T$ be a [[Definition:subset|subset]] of $S$.
Let $U$ be the [[Definition:Upper Closure of Subset|upper closure]] of $T$.
Then $U$ is an [[Definition:Upper Section|upper section]]. | Let $a \in U$.
Let $b \in S$ with $a \preceq b$.
By the definition of [[Definition:Upper Closure of Subset|upper closure]], there is a $t \in T$ such that $t \preceq a$.
By [[Definition:Transitive Relation|transitivity]], $t \preceq b$.
Thus, again by the definition of [[Definition:Upper Closure of Subset|upper clo... | Upper Closure is Upper Section | https://proofwiki.org/wiki/Upper_Closure_is_Upper_Section | https://proofwiki.org/wiki/Upper_Closure_is_Upper_Section | [
"Upper Sections",
"Upper Closures"
] | [
"Definition:Ordered Set",
"Definition:subset",
"Definition:Upper Closure/Set",
"Definition:Upper Section"
] | [
"Definition:Upper Closure/Set",
"Definition:Transitive Relation",
"Definition:Upper Closure/Set",
"Definition:Upper Section"
] |
proofwiki-7379 | Open Ray is Open in GO-Space/Definition 1 | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space.
Let $p \in S$.
Then:
:$p^\prec$ and $p^\succ$ are $\tau$-open
where:
:$p^\prec$ is the strict lower closure of $p$
:$p^\succ$ is the strict upper closure of $p$. | We will prove that $U = p^\succ$ is $\tau$-open.
That $p^\prec$ is $\tau$-open will follow by duality.
Let $u \in U$.
Since $p \notin U$, $p \ne u$.
By definition of GO-space, $\tau$ is Hausdorff.
From $T_2$ Space is $T_1$ Space, $\tau$ is $T_1$.
Thus by definition of GO-space, there is an open, order-convex set $M$ su... | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space|generalized ordered space]].
Let $p \in S$.
Then:
:$p^\prec$ and $p^\succ$ are [[Definition:Open Set (Topology)|$\tau$-open]]
where:
:$p^\prec$ is the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $p$
:$p^\succ$ is... | We will prove that $U = p^\succ$ is [[Definition:Open Set (Topology)|$\tau$-open]].
That $p^\prec$ is [[Definition:Open Set (Topology)|$\tau$-open]] will follow by duality.
Let $u \in U$.
Since $p \notin U$, $p \ne u$.
By definition of [[Definition:Generalized Ordered Space|GO-space]], $\tau$ is [[Definition:Hausdo... | Open Ray is Open in GO-Space/Definition 1 | https://proofwiki.org/wiki/Open_Ray_is_Open_in_GO-Space/Definition_1 | https://proofwiki.org/wiki/Open_Ray_is_Open_in_GO-Space/Definition_1 | [
"Generalized Ordered Spaces"
] | [
"Definition:Generalized Ordered Space",
"Definition:Open Set/Topology",
"Definition:Strict Lower Closure/Element",
"Definition:Strict Upper Closure/Element"
] | [
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Generalized Ordered Space",
"Definition:T2 Space",
"T2 Space is T1",
"Definition:Generalized Ordered Space",
"Definition:Order-Convex Set",
"Definition:Order-Convex Set",
"Set is Open iff Neighborhood of all its Points"
] |
proofwiki-7380 | Upper and Lower Closures of Open Set in GO-Space are Open | Let $\struct {X, \preceq, \tau}$ be a Generalized Ordered Space/Definition 1.
Let $A$ be open in $X$.
Then the upper and lower closures of $A$ are open. | We will show that the upper closure $U$ of $A$ is open.
The lower closure can be proven open by the same method.
By the definition of upper closure:
:$U = \set {u \in X: \exists a \in A: a \preceq u}$
But then:
{{begin-eqn}}
{{eqn | l = U
| r = \set {u \in X: \paren {u \in A} \lor \paren {\exists a \in A: a \prec... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space/Definition 1|Generalized Ordered Space/Definition 1]].
Let $A$ be open in $X$.
Then the upper and lower closures of $A$ are open. | We will show that the upper closure $U$ of $A$ is open.
The lower closure can be proven open by the same method.
By the definition of upper closure:
:$U = \set {u \in X: \exists a \in A: a \preceq u}$
But then:
{{begin-eqn}}
{{eqn | l = U
| r = \set {u \in X: \paren {u \in A} \lor \paren {\exists a \in A: a \p... | Upper and Lower Closures of Open Set in GO-Space are Open | https://proofwiki.org/wiki/Upper_and_Lower_Closures_of_Open_Set_in_GO-Space_are_Open | https://proofwiki.org/wiki/Upper_and_Lower_Closures_of_Open_Set_in_GO-Space_are_Open | [
"Topology",
"Total Orderings"
] | [
"Definition:Generalized Ordered Space/Definition 1"
] | [
"Definition:Upper Closure/Element",
"Open Ray is Open in GO-Space/Definition 1",
"Category:Topology",
"Category:Total Orderings"
] |
proofwiki-7381 | Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3 | Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 1:
{{:Definition:Generalized Ordered Space/Definition 1}}
Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 3:
{{:Definition:Generalized Ordered Space/Definition 3}} | Let $\BB$ be a basis for $\tau$ consisting of order-convex sets.
Let:
:$\SS = \set {U^\succeq: U \in \BB} \cup \set {U^\preceq: U \in \BB}$
where $U^\succeq$ and $U^\preceq$ denote the upper closure and lower closure respectively of $U$.
By Upper Closure is Upper Section and Lower Closure is Lower Section, the elements... | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space/Definition 1|generalized ordered space by Definition 1]]:
{{:Definition:Generalized Ordered Space/Definition 1}}
Then $\struct {S, \preceq, \tau}$ is a [[Definition:Generalized Ordered Space/Definition 3|generalized ordered space by Definiti... | Let $\BB$ be a [[Definition:Basis (Topology)|basis]] for $\tau$ consisting of [[Definition:Order-Convex Set|order-convex sets]].
Let:
:$\SS = \set {U^\succeq: U \in \BB} \cup \set {U^\preceq: U \in \BB}$
where $U^\succeq$ and $U^\preceq$ denote the [[Definition:Upper Closure of Subset|upper closure]] and [[Definition:... | Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space/Definition_1_implies_Definition_3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Generalized_Ordered_Space/Definition_1_implies_Definition_3 | [
"Equivalence of Definitions of Generalized Ordered Space"
] | [
"Definition:Generalized Ordered Space/Definition 1",
"Definition:Generalized Ordered Space/Definition 3"
] | [
"Definition:Basis (Topology)",
"Definition:Order-Convex Set",
"Definition:Upper Closure/Set",
"Definition:Lower Closure/Set",
"Upper Closure is Upper Section",
"Lower Closure is Lower Section",
"Definition:Element",
"Definition:Upper Section",
"Definition:Lower Section",
"Definition:Sub-Basis",
... |
proofwiki-7382 | Upper Section is Order-Convex | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be an upper section.
Then $T$ is order-convex in $S$. | Let $a, c \in T$.
Let $b \in S$.
Let $a \preceq b \preceq c$.
Since:
:$a \in T$
:$a \preceq b$
:$T$ is an upper section
it follows that:
:$b \in T$
This holds for all such $a$, $b$, and $c$.
Therefore, by definition, $T$ is order-convex in $S$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$ be an [[Definition:Upper Section|upper section]].
Then $T$ is [[Definition:Order-Convex Set|order-convex]] in $S$. | Let $a, c \in T$.
Let $b \in S$.
Let $a \preceq b \preceq c$.
Since:
:$a \in T$
:$a \preceq b$
:$T$ is an [[Definition:Upper Section|upper section]]
it follows that:
:$b \in T$
This holds for all such $a$, $b$, and $c$.
Therefore, by definition, $T$ is [[Definition:Order-Convex Set|order-convex]] in $S$.
{{qed}} | Upper Section is Order-Convex | https://proofwiki.org/wiki/Upper_Section_is_Order-Convex | https://proofwiki.org/wiki/Upper_Section_is_Order-Convex | [
"Upper Sections",
"Order-Convex Sets"
] | [
"Definition:Ordered Set",
"Definition:Upper Section",
"Definition:Order-Convex Set"
] | [
"Definition:Upper Section",
"Definition:Order-Convex Set"
] |
proofwiki-7383 | Lower Section is Order-Convex | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a lower section.
Then $T$ is order-convex in $S$. | Let $a, c \in T$.
Let $b \in S$.
Let $a \preceq b \preceq c$.
Since:
:$c \in T$
:$b \preceq c$
:$T$ is a lower section
it follows that:
:$b \in T$
This holds for all such $a$, $b$, and $c$.
Hence, by definition, $T$ is order-convex in $S$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$ be a [[Definition:Lower Section|lower section]].
Then $T$ is [[Definition:Order-Convex Set|order-|convex]] in $S$. | Let $a, c \in T$.
Let $b \in S$.
Let $a \preceq b \preceq c$.
Since:
:$c \in T$
:$b \preceq c$
:$T$ is a [[Definition:Lower Section|lower section]]
it follows that:
:$b \in T$
This holds for all such $a$, $b$, and $c$.
Hence, by definition, $T$ is [[Definition:Order-Convex Set|order-convex]] in $S$.
{{qed}} | Lower Section is Order-Convex | https://proofwiki.org/wiki/Lower_Section_is_Order-Convex | https://proofwiki.org/wiki/Lower_Section_is_Order-Convex | [
"Lower Sections",
"Order-Convex Sets"
] | [
"Definition:Ordered Set",
"Definition:Lower Section",
"Definition:Order-Convex Set"
] | [
"Definition:Lower Section",
"Definition:Order-Convex Set"
] |
proofwiki-7384 | GO-Space Embeds Densely into Linearly Ordered Space | Let $\struct {Y, \preceq, \tau}$ be a generalized ordered space (GO-space) by Definition 3.
That is:
:let $\struct {Y, \tau}$ be a Hausdorff space
and:
:let $\tau$ have a sub-basis consisting of upper sections and lower sections relative to $\preceq$.
Then $\struct {Y, \preceq, \tau}$ is a GO-space by Definition 2.
Tha... | <section notitle="True" name="proof">
Let $X$ be the disjoint union of $Y$ with the set of all lower sections $L$ in $Y$ such that $L$ and $Y \setminus L$ are open and nonempty and either:
:$L$ has a maximum, and $Y\setminus L$ does not have a minimum, or
:$Y \setminus L$ has a minimum, and $L$ does not have a maximum... | Let $\struct {Y, \preceq, \tau}$ be a [[Definition:Generalized Ordered Space/Definition 3|generalized ordered space (GO-space) by Definition 3]].
That is:
:let $\struct {Y, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]]
and:
:let $\tau$ have a [[Definition:Sub-Basis|sub-basis]] consisting of [[Definition:U... | <section notitle="True" name="proof">
Let $X$ be the disjoint union of $Y$ with the set of all [[Definition:Lower Section|lower sections]] $L$ in $Y$ such that $L$ and $Y \setminus L$ are open and nonempty and either:
:$L$ has a maximum, and $Y\setminus L$ does not have a minimum, or
:$Y \setminus L$ has a minimum, an... | GO-Space Embeds Densely into Linearly Ordered Space | https://proofwiki.org/wiki/GO-Space_Embeds_Densely_into_Linearly_Ordered_Space | https://proofwiki.org/wiki/GO-Space_Embeds_Densely_into_Linearly_Ordered_Space | [
"Generalized Ordered Spaces",
"Linearly Ordered Spaces"
] | [
"Definition:Generalized Ordered Space/Definition 3",
"Definition:T2 Space",
"Definition:Sub-Basis",
"Definition:Upper Section",
"Definition:Lower Section",
"Definition:Generalized Ordered Space/Definition 2",
"Definition:Linearly Ordered Space",
"Definition:Mapping",
"Definition:Order Embedding",
... | [
"Definition:Lower Section",
"Definition:Lower Section",
"Union of Total Ordering with Lower Sections is Total Ordering",
"Restriction of Total Ordering is Total Ordering",
"Definition:Total Ordering",
"Definition:Order Embedding",
"Definition:Order Topology",
"Definition:Lower Section",
"Definition:... |
proofwiki-7385 | Upper Section with no Minimal Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $U \subseteq S$.
Then:
:$U$ is an upper section in $S$ with no minimal element
{{iff}}:
:$\ds U = \bigcup \set {u^\succ: u \in U}$
where $u^\succ$ is the strict upper closure of $u$. | === Forward implication ===
Let $U$ be an upper section in $S$ with no minimal element.
Then by the definition of upper section:
:$\ds \bigcup \set {u^\succ: u \in U} \subseteq U$
Let $x \in U$.
Since $U$ has no minimal element, $x$ is not minimal.
Thus there is a $u \in U$ such that $u \prec x$.
Then $x \in u^\succ$, ... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $U \subseteq S$.
Then:
:$U$ is an [[Definition:Upper Section|upper section]] in $S$ with no [[Definition:Minimal Element|minimal element]]
{{iff}}:
:$\ds U = \bigcup \set {u^\succ: u \in U}$
where $u^\succ$ is the [[Definition:Strict Upper ... | === Forward implication ===
Let $U$ be an [[Definition:Upper Section|upper section]] in $S$ with no [[Definition:Minimal Element|minimal element]].
Then by the definition of [[Definition:Upper Section|upper section]]:
:$\ds \bigcup \set {u^\succ: u \in U} \subseteq U$
Let $x \in U$.
Since $U$ has no [[Definition:Mi... | Upper Section with no Minimal Element | https://proofwiki.org/wiki/Upper_Section_with_no_Minimal_Element | https://proofwiki.org/wiki/Upper_Section_with_no_Minimal_Element | [
"Upper Sections"
] | [
"Definition:Ordered Set",
"Definition:Upper Section",
"Definition:Minimal/Element",
"Definition:Strict Upper Closure/Element"
] | [
"Definition:Upper Section",
"Definition:Minimal/Element",
"Definition:Upper Section",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Set Equality/Definition 2",
"Definition:Upper Section",
"Definition:Minimal/Element",
"Definition:Minimal/Element"
] |
proofwiki-7386 | Lower Sections in Totally Ordered Set form Chain | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\LL$ be a set of lower sections in $S$.
Then $\LL$ is a chain.
That is, $\LL$ is totally ordered by $\subseteq$. | Let $L, M \in \LL$.
Suppose that $M \nsubseteq L$.
Then:
:$\exists x \in M: x \notin L$
Let $y \in L$.
Then since $\preceq$ is a total ordering, $x \preceq y$ or $y \preceq x$.
If $x \preceq y$, then since $L$ is a lower section: $x \in L$, a contradiction.
Thus $y \preceq x$.
Since $M$ is a lower section, $y \in M$.
S... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\LL$ be a [[Definition:Set|set]] of [[Definition:Lower Section|lower sections]] in $S$.
Then $\LL$ is a [[Definition:Chain of Sets|chain]].
That is, $\LL$ is [[Definition:Totally Ordered Set|totally ordered]] by $\subseteq$... | Let $L, M \in \LL$.
Suppose that $M \nsubseteq L$.
Then:
:$\exists x \in M: x \notin L$
Let $y \in L$.
Then since $\preceq$ is a [[Definition:Total Ordering|total ordering]], $x \preceq y$ or $y \preceq x$.
If $x \preceq y$, then since $L$ is a [[Definition:Lower Section|lower section]]: $x \in L$, a contradicti... | Lower Sections in Totally Ordered Set form Chain | https://proofwiki.org/wiki/Lower_Sections_in_Totally_Ordered_Set_form_Chain | https://proofwiki.org/wiki/Lower_Sections_in_Totally_Ordered_Set_form_Chain | [
"Lower Sections",
"Total Orderings",
"Chains (Order Theory)"
] | [
"Definition:Totally Ordered Set",
"Definition:Set",
"Definition:Lower Section",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Totally Ordered Set"
] | [
"Definition:Total Ordering",
"Definition:Lower Section",
"Definition:Lower Section",
"Definition:Chain (Order Theory)/Subset Relation",
"Category:Lower Sections",
"Category:Total Orderings",
"Category:Chains (Order Theory)"
] |
proofwiki-7387 | Exclusive Or is Self-Inverse | :$\paren {p \oplus q} \oplus q \dashv \vdash p$
where $\oplus$ denotes the exclusive or operator. | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective on the {{LHS}} match those for $p$ on the {{RHS}} for all boolean interpretations:
$\begin{array}{|ccccc||c|} \hline
(p & \oplus & q) & \oplus & q & p \\
\hline
\F & \F & \F & \F & \F & \F \\
\F & \T & \T & \F ... | :$\paren {p \oplus q} \oplus q \dashv \vdash p$
where $\oplus$ denotes the [[Definition:Exclusive Or|exclusive or operator]]. | 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]] on the {{LHS}} match those for $p$ on the {{RHS}} for all [[Definition:Boolean Interpretation|boolean interpretations]]:
$\b... | Exclusive Or is Self-Inverse | https://proofwiki.org/wiki/Exclusive_Or_is_Self-Inverse | https://proofwiki.org/wiki/Exclusive_Or_is_Self-Inverse | [
"Exclusive Or"
] | [
"Definition:Exclusive Or"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation"
] |
proofwiki-7388 | Conjunction has no Inverse | Let $\land$ denote the conjunction operation of propositional logic.
Then there exists no binary logical connective $\circ$ such that:
:$(1): \quad \forall p, q \in \set {T, F}: \left({p \land q}\right) \circ q = p$ | This will be proven by contradiction.
Let such an operation $\circ$ exist.
Let $f^\circ: \mathbb B^2 \to \mathbb B$ be the associated truth function.
Suppose now that $q = F$, while $p$ remains unspecified.
Then:
:$p \land q = \map {f^\land} {p, F} = F$
where $f^\land$ is the truth function of conjunction.
It does not ... | Let $\land$ denote the [[Definition:Conjunction|conjunction operation]] of [[Definition:Propositional Logic|propositional logic]].
Then there exists no [[Definition:Binary Logical Connective|binary logical connective]] $\circ$ such that:
:$(1): \quad \forall p, q \in \set {T, F}: \left({p \land q}\right) \circ q = p... | This will be [[Proof by Contradiction|proven by contradiction]].
Let such an operation $\circ$ exist.
Let $f^\circ: \mathbb B^2 \to \mathbb B$ be the associated [[Definition:Truth Function|truth function]].
Suppose now that $q = F$, while $p$ remains unspecified.
Then:
:$p \land q = \map {f^\land} {p, F} = F$
wh... | Conjunction has no Inverse | https://proofwiki.org/wiki/Conjunction_has_no_Inverse | https://proofwiki.org/wiki/Conjunction_has_no_Inverse | [
"Conjunction"
] | [
"Definition:Conjunction",
"Definition:Propositional Logic",
"Definition:Logical Connective/Binary"
] | [
"Proof by Contradiction",
"Definition:Truth Function",
"Definition:Truth Function",
"Definition:Conjunction",
"Category:Conjunction"
] |
proofwiki-7389 | Disjunction has no Inverse | Let $\lor$ denote the disjunction operation of propositional logic.
Then there exists no binary logical connective $\circ$ such that:
:$(1): \quad \forall p, q \in \left\{{T, F}\right\}: \left({p \lor q}\right) \circ q = p$ | Let $q$ be true.
Then $p \lor q = T$, whatever truth value $p$ holds.
In this case $(1)$ simplifies to:
:$(2): \quad \forall p \in \left\{{T, F}\right\}: T \circ T = p$
Either $T \circ T = T$ or $T \circ T = F$, but not both.
If $p = F$, then it must be that $T \circ T = F$.
If $p = T$, then it must be that $T \circ T ... | Let $\lor$ denote the [[Definition:Disjunction|disjunction operation]] of [[Definition:Propositional Logic|propositional logic]].
Then there exists no [[Definition:Binary Logical Connective|binary logical connective]] $\circ$ such that:
:$(1): \quad \forall p, q \in \left\{{T, F}\right\}: \left({p \lor q}\right) \ci... | Let $q$ be [[Definition:True|true]].
Then $p \lor q = T$, whatever [[Definition:Truth Value|truth value]] $p$ holds.
In this case $(1)$ simplifies to:
:$(2): \quad \forall p \in \left\{{T, F}\right\}: T \circ T = p$
Either $T \circ T = T$ or $T \circ T = F$, but not both.
If $p = F$, then it must be that $T \circ T... | Disjunction has no Inverse | https://proofwiki.org/wiki/Disjunction_has_no_Inverse | https://proofwiki.org/wiki/Disjunction_has_no_Inverse | [
"Disjunction"
] | [
"Definition:Disjunction",
"Definition:Propositional Logic",
"Definition:Logical Connective/Binary"
] | [
"Definition:True",
"Definition:Truth Value",
"Category:Disjunction"
] |
proofwiki-7390 | Binary Logical Connectives with Inverse | Let $\circ$ be a binary logical connective.
Then there exists another binary logical connective $*$ such that:
:$\forall p, q \in \set {\F, \T}: \paren {p \circ q} * q \dashv \vdash p \dashv \vdash q * \paren {p \circ q}$
{{iff}} $\circ$ is either:
:$(1): \quad$ the exclusive or operator
or:
:$(2): \quad$ the biconditi... | === Necessary Condition ===
Let $\circ$ be a binary logical connective such that there exists $*$ such that:
:$\paren {p \circ q} * q \dashv \vdash p$
That is, by definition (and minor abuse of notation):
:$\forall p, q \in \set {\F, \T}: \paren {p \circ q} * q = p$
For reference purposes, let us list from Binary Truth... | Let $\circ$ be a [[Definition:Binary Logical Connective|binary logical connective]].
Then there exists another [[Definition:Binary Logical Connective|binary logical connective]] $*$ such that:
:$\forall p, q \in \set {\F, \T}: \paren {p \circ q} * q \dashv \vdash p \dashv \vdash q * \paren {p \circ q}$
{{iff}} $\circ$... | === Necessary Condition ===
Let $\circ$ be a [[Definition:Binary Logical Connective|binary logical connective]] such that there exists $*$ such that:
:$\paren {p \circ q} * q \dashv \vdash p$
That is, by definition (and minor abuse of notation):
:$\forall p, q \in \set {\F, \T}: \paren {p \circ q} * q = p$
For refe... | Binary Logical Connectives with Inverse | https://proofwiki.org/wiki/Binary_Logical_Connectives_with_Inverse | https://proofwiki.org/wiki/Binary_Logical_Connectives_with_Inverse | [
"Exclusive Or",
"Biconditional",
"Propositional Logic"
] | [
"Definition:Logical Connective/Binary",
"Definition:Logical Connective/Binary",
"Definition:Exclusive Or",
"Definition:Biconditional",
"Definition:Truth Function",
"Definition:Inverse Operation",
"Definition:Exclusive Or",
"Definition:Biconditional"
] | [
"Definition:Logical Connective/Binary",
"Binary Truth Functions",
"Definition:Logical Connective/Binary",
"Definition:Inverse Operation",
"Definition:Logical Connective/Binary",
"Definition:Inverse Operation",
"Definition:Exclusive Or",
"Definition:Biconditional",
"Exclusive Or is Self-Inverse",
"... |
proofwiki-7391 | Biconditional is Self-Inverse | :$\paren {p \iff q} \iff q \dashv \vdash p$
where $\iff$ denotes the biconditional operator. | We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connective on the {{LHS}} match those for $p$ on the {{RHS}} for all boolean interpretations:
:<nowiki>$\begin{array}{|ccccc||c|} \hline
(p & \iff & q) & \iff & q & p \\
\hline
\F & \T & \F & \F & \F & \F \\
\F & \F & \T ... | :$\paren {p \iff q} \iff q \dashv \vdash p$
where $\iff$ denotes the [[Definition:Biconditional|biconditional operator]]. | 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]] on the {{LHS}} match those for $p$ on the {{RHS}} for all [[Definition:Boolean Interpretation|boolean interpretations]]:
:<n... | Biconditional is Self-Inverse | https://proofwiki.org/wiki/Biconditional_is_Self-Inverse | https://proofwiki.org/wiki/Biconditional_is_Self-Inverse | [
"Biconditional",
"Truth Table Proofs"
] | [
"Definition:Biconditional"
] | [
"Method of Truth Tables",
"Definition:Truth Value",
"Definition:Main Connective/Propositional Logic",
"Definition:Boolean Interpretation",
"Category:Biconditional",
"Category:Truth Table Proofs"
] |
proofwiki-7392 | Finite Chain is Order-Isomorphic to Finite Ordinal | Let $\struct {S, \preceq}$ be an ordered set.
Let $C$ be a finite chain in $S$.
Then for some finite ordinal $\mathbf n$:
:$\struct {C, {\preceq \restriction_C} }$ is order-isomorphic to $\mathbf n$.
That is:
:$\struct {C, {\preceq \restriction_C} }$ is order-isomorphic to $\N_n$
where $\N_n$ is the initial segment of ... | By the definition of finite set:
:there exists an $n \in \N$ such that there exists a bijection $f: C \to \N_n$.
This $n$ is unique by Equality of Natural Numbers and Set Equivalence behaves like Equivalence Relation.
Define a mapping $g: \N_n \to C$ recursively as:
:$\map g 0 = \min C$
:$\map g {k + 1} = \map \min {C ... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $C$ be a [[Definition:Finite Set|finite]] [[Definition:Chain (Order Theory)|chain]] in $S$.
Then for some finite [[Definition:Ordinal|ordinal]] $\mathbf n$:
:$\struct {C, {\preceq \restriction_C} }$ is [[Definition:Order Isomorphism|order-i... | By the definition of [[Definition:Finite Set|finite set]]:
:there exists an $n \in \N$ such that there exists a [[Definition:Bijection|bijection]] $f: C \to \N_n$.
This $n$ is [[Definition:Unique|unique]] by [[Equality of Natural Numbers]] and [[Set Equivalence behaves like Equivalence Relation]].
Define a [[Definit... | Finite Chain is Order-Isomorphic to Finite Ordinal | https://proofwiki.org/wiki/Finite_Chain_is_Order-Isomorphic_to_Finite_Ordinal | https://proofwiki.org/wiki/Finite_Chain_is_Order-Isomorphic_to_Finite_Ordinal | [
"Total Orderings",
"Ordinals",
"Chains (Order Theory)"
] | [
"Definition:Ordered Set",
"Definition:Finite Set",
"Definition:Chain (Order Theory)",
"Definition:Ordinal",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism",
"Definition:Initial Segment of Natural Numbers/Zero-Based"
] | [
"Definition:Finite Set",
"Definition:Bijection",
"Definition:Unique",
"Equality of Natural Numbers",
"Set Equivalence behaves like Equivalence Relation",
"Definition:Mapping",
"Principle of Recursive Definition"
] |
proofwiki-7393 | Complete Linearly Ordered Space is Compact | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $\struct {X, \preceq}$ be a complete lattice.
Then $\struct {X, \tau}$ is compact. | By Space is Compact iff exists Basis such that Every Cover has Finite Subcover, it is sufficient to prove that an open cover of $X$ consisting of open intervals and rays has a finite subcover.
Let $\AA$ be an open cover of $X$ consisting of open rays and open intervals.
Let $m = \inf X$.
This infimum exists because $\s... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $\struct {X, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Then $\struct {X, \tau}$ is [[Definition:Compact Topological Space|compact]]. | By [[Space is Compact iff exists Basis such that Every Cover has Finite Subcover]], it is sufficient to prove that an open cover of $X$ consisting of open intervals and rays has a finite subcover.
Let $\AA$ be an open cover of $X$ consisting of open rays and open intervals.
Let $m = \inf X$.
This infimum exists beca... | Complete Linearly Ordered Space is Compact | https://proofwiki.org/wiki/Complete_Linearly_Ordered_Space_is_Compact | https://proofwiki.org/wiki/Complete_Linearly_Ordered_Space_is_Compact | [
"Linearly Ordered Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Complete Lattice",
"Definition:Compact Topological Space"
] | [
"Space is Compact iff Every Cover from Basis has Finite Subcover"
] |
proofwiki-7394 | Infinite Sequence Property of Well-Founded Relation/Reverse Implication | Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be such that there exists no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
Then $\RR$ is a well-founded relation. | Suppose $\RR$ is not a well-founded relation.
So by definition there exists a non-empty subset $T$ of $S$ which has no minimal element.
Let $a \in T$.
Since $a$ is not minimal in $T$, we can find $b \in T: \paren {b \mathrel \RR a} \text { and } \paren {b \ne a}$.
This holds for all $a \in T$.
Hence the restriction $\R... | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]].
Let $\RR$ be such that there exists no [[Definition:Infinite Sequence|infinite sequence]] $\sequence {a_n}$ of [[Definition:Element|elements]] of $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \p... | Suppose $\RR$ is not a [[Definition:Well-Founded Relation|well-founded relation]].
So by definition there exists a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $T$ of $S$ which has no [[Definition:Minimal Element|minimal element]].
Let $a \in T$.
Since $a$ is not [[Definition:Minimal Element|m... | Infinite Sequence Property of Well-Founded Relation/Reverse Implication/Proof 1 | https://proofwiki.org/wiki/Infinite_Sequence_Property_of_Well-Founded_Relation/Reverse_Implication | https://proofwiki.org/wiki/Infinite_Sequence_Property_of_Well-Founded_Relation/Reverse_Implication/Proof_1 | [
"Infinite Sequence Property of Well-Founded Relation"
] | [
"Definition:Relational Structure",
"Definition:Sequence/Infinite Sequence",
"Definition:Element",
"Definition:Well-Founded Relation"
] | [
"Definition:Well-Founded Relation",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Restriction/Relation",
"Definition:Right-Total Relation",
"Definition:Endorelation",
"Axiom:Axiom of Dependent Choice/Right-Total",
"Definiti... |
proofwiki-7395 | Infinite Sequence Property of Well-Founded Relation/Reverse Implication | Let $\struct {S, \RR}$ be a relational structure.
Let $\RR$ be such that there exists no infinite sequence $\sequence {a_n}$ of elements of $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \paren {a_{n + 1} \ne a_n}$
Then $\RR$ is a well-founded relation. | Suppose $\RR$ is not a well-founded relation.
Hence there exists $T \subseteq S$ such that $T$ has no minimal element under $\RR$.
Let $a_0 \in T$.
We have that $a_0$ is not minimal in $T$.
So:
:$\exists a_1 \in T: \paren {a_1 \mathrel \RR a_0} \text { and } a_1 \ne a_0$
Similarly, $a_1$ is not minimal in $T$.
So:
:$\e... | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]].
Let $\RR$ be such that there exists no [[Definition:Infinite Sequence|infinite sequence]] $\sequence {a_n}$ of [[Definition:Element|elements]] of $S$ such that:
:$\forall n \in \N: \paren {a_{n + 1} \mathrel \RR a_n} \text { and } \p... | Suppose $\RR$ is not a [[Definition:Well-Founded Relation|well-founded relation]].
Hence there exists $T \subseteq S$ such that $T$ has no [[Definition:Minimal Element|minimal element]] under $\RR$.
Let $a_0 \in T$.
We have that $a_0$ is not [[Definition:Minimal Element|minimal]] in $T$.
So:
:$\exists a_1 \in T: \... | Infinite Sequence Property of Well-Founded Relation/Reverse Implication/Proof 2 | https://proofwiki.org/wiki/Infinite_Sequence_Property_of_Well-Founded_Relation/Reverse_Implication | https://proofwiki.org/wiki/Infinite_Sequence_Property_of_Well-Founded_Relation/Reverse_Implication/Proof_2 | [
"Infinite Sequence Property of Well-Founded Relation"
] | [
"Definition:Relational Structure",
"Definition:Sequence/Infinite Sequence",
"Definition:Element",
"Definition:Well-Founded Relation"
] | [
"Definition:Well-Founded Relation",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Definition:Minimal/Element",
"Axiom:Axiom of Dependent Choice/Right-Total",
"Definition:Minimal/Element",
"Definition:Right-Total Relation",
"Axiom:Axiom of Dependent Choice/Right-Total",
"Definition:Seq... |
proofwiki-7396 | Inversion Mapping on Ordered Group is Dual Order-Isomorphism | Let $\struct {G, \circ, \preceq}$ be an ordered group.
Let $\iota: G \to G$ be the inversion mapping, defined by $\map \phi x = x^{-1}$.
Then $\iota$ is a dual order-isomorphism. | By Inversion Mapping is Involution and Involution is Permutation, $\iota$ is a permutation and so by definition bijective.
Let $x, y \in G$ such that $x \prec y$.
Then $y^{-1} \prec x^{-1}$ by Inversion Mapping Reverses Ordering in Ordered Group.
Thus $\map \iota y \prec \map \iota x$.
Since this holds for all $x$ and ... | Let $\struct {G, \circ, \preceq}$ be an [[Definition:Ordered Group|ordered group]].
Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]], defined by $\map \phi x = x^{-1}$.
Then $\iota$ is a [[Definition:Dual Isomorphism (Order Theory)|dual order-isomorphism]]. | By [[Inversion Mapping is Involution]] and [[Involution is Permutation]], $\iota$ is a [[Definition:Permutation|permutation]] and so by definition [[Definition:Bijection|bijective]].
Let $x, y \in G$ such that $x \prec y$.
Then $y^{-1} \prec x^{-1}$ by [[Inversion Mapping Reverses Ordering in Ordered Group]].
Thus $... | Inversion Mapping on Ordered Group is Dual Order-Isomorphism | https://proofwiki.org/wiki/Inversion_Mapping_on_Ordered_Group_is_Dual_Order-Isomorphism | https://proofwiki.org/wiki/Inversion_Mapping_on_Ordered_Group_is_Dual_Order-Isomorphism | [
"Ordered Groups",
"Inversion Mappings"
] | [
"Definition:Ordered Group",
"Definition:Inversion Mapping",
"Definition:Dual Isomorphism (Order Theory)"
] | [
"Inversion Mapping is Involution",
"Involution is Permutation",
"Definition:Permutation",
"Definition:Bijection",
"Inversion Mapping Reverses Ordering in Ordered Group",
"Definition:Strictly Decreasing/Mapping",
"Inverse of Group Inverse",
"Definition:Dual Isomorphism (Order Theory)",
"Category:Orde... |
proofwiki-7397 | Ordering Compatible with Group Operation is Strongly Compatible/Corollary/Proof 1 | Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $\prec$ be the reflexive reduction of $\preceq$.
Let $x, y \in G$.
Then the following equivalences hold:
{{:Ordering Compatible with Group Operation is Strongly Compatible/Corollary}} | By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by {{Corollary|Relation Compatible with Group Operation is Strongly Compatible}}, we obtain the first four results.
By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is compatible wi... | Let $\struct {G, \circ, \preccurlyeq}$ be an [[Definition:Ordered Group|ordered group]] with [[Definition:Identity Element|identity]] $e$.
Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preceq$.
Let $x, y \in G$.
Then the following equivalences hold:
{{:Ordering Compatible with Group... | By the definition of an [[Definition:Ordered Group|ordered group]], $\preccurlyeq$ is a [[Definition:Relation Compatible with Operation|relation compatible]] with $\circ$.
Thus by {{Corollary|Relation Compatible with Group Operation is Strongly Compatible}}, we obtain the first four results.
By [[Reflexive Reduction... | Ordering Compatible with Group Operation is Strongly Compatible/Corollary/Proof 1 | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary/Proof_1 | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary/Proof_1 | [
"Ordering Compatible with Group Operation is Strongly Compatible"
] | [
"Definition:Ordered Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Reflexive Reduction"
] | [
"Definition:Ordered Group",
"Definition:Relation Compatible with Operation",
"Reflexive Reduction of Relation Compatible with Group Operation is Compatible",
"Definition:Relation Compatible with Operation"
] |
proofwiki-7398 | Ordering Compatible with Group Operation is Strongly Compatible/Corollary/Proof 2 | Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $\prec$ be the reflexive reduction of $\preccurlyeq$.
Let $x, y \in G$.
Then the following equivalences hold:
{{:Ordering Compatible with Group Operation is Strongly Compatible/Corollary}} | Each result follows from Ordering Compatible with Group Operation is Strongly Compatible.
For example, by Ordering Compatible with Group Operation is Strongly Compatible:
:$x \preccurlyeq y \iff x \circ x^{-1} \preccurlyeq y \circ x^{-1}$
Since $x \circ x^{-1} = e$:
:$x \preccurlyeq y \iff e \preccurlyeq y \circ x^{-1}... | Let $\struct {G, \circ, \preccurlyeq}$ be an [[Definition:Ordered Group|ordered group]] with [[Definition:Identity Element|identity]] $e$.
Let $\prec$ be the [[Definition:Reflexive Reduction|reflexive reduction]] of $\preccurlyeq$.
Let $x, y \in G$.
Then the following equivalences hold:
{{:Ordering Compatible with ... | Each result follows from [[Ordering Compatible with Group Operation is Strongly Compatible]].
For example, by [[Ordering Compatible with Group Operation is Strongly Compatible]]:
:$x \preccurlyeq y \iff x \circ x^{-1} \preccurlyeq y \circ x^{-1}$
Since $x \circ x^{-1} = e$:
:$x \preccurlyeq y \iff e \preccurlyeq y ... | Ordering Compatible with Group Operation is Strongly Compatible/Corollary/Proof 2 | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary/Proof_2 | https://proofwiki.org/wiki/Ordering_Compatible_with_Group_Operation_is_Strongly_Compatible/Corollary/Proof_2 | [
"Ordering Compatible with Group Operation is Strongly Compatible"
] | [
"Definition:Ordered Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Reflexive Reduction"
] | [
"Ordering Compatible with Group Operation is Strongly Compatible",
"Ordering Compatible with Group Operation is Strongly Compatible"
] |
proofwiki-7399 | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 1 | Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $x \in G$.
Then the following equivalences hold:
{{:Inversion Mapping Reverses Ordering in Ordered Group/Corollary}} | By Inversion Mapping Reverses Ordering in Ordered Group:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e^{-1} \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e^{-1}
}}
{{eqn | l = x \prec e
| o = \iff
... | Let $\struct {G, \circ, \preccurlyeq}$ be an [[Definition:Ordered Group|ordered group]] with [[Definition:Identity Element|identity]] $e$.
Let $x \in G$.
Then the following equivalences hold:
{{:Inversion Mapping Reverses Ordering in Ordered Group/Corollary}} | By [[Inversion Mapping Reverses Ordering in Ordered Group]]:
{{begin-eqn}}
{{eqn | q = \forall x \in G
| l = x \preccurlyeq e
| o = \iff
| r = e^{-1} \preccurlyeq x^{-1}
}}
{{eqn | l = e \preccurlyeq x
| o = \iff
| r = x^{-1} \preccurlyeq e^{-1}
}}
{{eqn | l = x \prec e
| o = \iff
... | Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 1 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_1 | https://proofwiki.org/wiki/Inversion_Mapping_Reverses_Ordering_in_Ordered_Group/Corollary/Proof_1 | [
"Inversion Mapping Reverses Ordering in Ordered Group"
] | [
"Definition:Ordered Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Inversion Mapping Reverses Ordering in Ordered Group"
] |
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