hcju/proofwikiqa huggingface dataset:
Taking the group axioms in turn: === $\text G 0$: Closure === Let $f_{ab}, f_{cd} \in \map {\operatorname {Af}_1} \R$. Then: {{begin-eqn}} {{eqn | l = \map {f_{ab} \circ f_{cd} } x       | r = a \paren {c x + d} + b       | c =  }} {{eqn | r = a c x + a d + b       | c =  }} {{end-eqn}} By the field axioms, $a c \in \R_{\ne 0}$ and $a d + b \in \R$. Thus $f_{ab} \circ f_{cd} \in \map {\operatorname {Af}_1} \R$ and so $\map {\operatorname {Af}_1} \R$ is closed. {{qed|lemma}} === $\text G 1$: Associativity === From Composition of Mappings is Associative, it follows directly that $\circ$ is associative on $\map {\operatorname {Af}_1} \R$. {{qed|lemma}} === $\text G 2$: Identity === By Identity of Affine Group of One Dimension, $\map {\operatorname {Af}_1} \R$ has $f_{1, 0}$ as an identity element. {{qed|lemma}} === $\text G 3$: Inverses === By Inverse in Affine Group of One Dimension, every element $f_{a b}$ of $\map {\operatorname {Af}_1} \R$ has an inverse $f_{c d}$ where $c = \dfrac 1 a$ and $d = \dfrac {-b} a$. {{qed|lemma}} All the group axioms are thus seen to be fulfilled, and so $\map {\operatorname {Af}_1} \R$ is a group. {{qed}}
Let $f_{ab} \in \operatorname{Af}_1 \left({\R}\right)$. Then: {{begin-eqn}} {{eqn | l = f_{ab} \circ f_{1, 0} \left({x}\right)       | r = a \left({1 x + 0}\right) + b       | c =  }} {{eqn | r = a x + b       | c =  }} {{eqn | r = f_{ab}       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = f_{1, 0} \circ f_{a b} \left({x}\right)       | r = 1 \left({a x + b}\right) + 0       | c =  }} {{eqn | r = a x + b       | c =  }} {{eqn | r = f_{ab}       | c =  }} {{end-eqn}} Thus $f_{1, 0}$ is the identity element of $\operatorname{Af}_1 \left({\R}\right)$. {{qed}} Category:Affine Groups qg1cgk88mnkxd2maqjzdexhsrxhcwh9
{{begin-eqn}} {{eqn | l = a b c^{-1}       | r = b a c^{-1}       | c = as $a$ commutes with $b$ }} {{eqn | r = b c^{-1} a       | c = Commutation with Inverse in Monoid }} {{end-eqn}} {{qed}}
Let the number of vertices, edges and faces of $M$ be $V$, $E$ and $F$ respectively. From Möbius Strip has no Vertices: :$V = 0$ From Möbius Strip has 1 Edge: :$E = 1$ From Möbius Strip has 1 Face: :$F = 1$ By definition of the Euler characteristic: {{begin-eqn}} {{eqn | l = \map \chi M       | r = V - E + F       | c = Definition of Euler Characteristic }} {{eqn | r = 0 - 1 + 1       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}}
Let $g, h \in G$ and $x \in X$. We verify that $g * \paren {h * x} = \paren {g h} * x$: {{begin-eqn}} {{eqn | l = g * \paren {h * x}       | r = \map {\map \rho g} {\map {\map \rho h} x}       | c = Definition of $\phi$ }} {{eqn | r = \map {\paren {\map \rho g \circ \map \rho h} } x       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\map \rho {g h} } x       | c = $\rho$ is a homomorphism }} {{eqn | r = \paren {g h} * x       | c = Definition of $\phi$ }} {{end-eqn}} We verify that $e * x = x$. Let $I_X$ denote the identity mapping on $X$. {{begin-eqn}} {{eqn | l = e*x       | r = \map {\map \rho e} x       | c = Definition of $\phi$ }} {{eqn | r = \map {I_X} x       | c = Group Homomorphism Preserves Identity, Set of all Self-Maps is Monoid }} {{eqn | r = x       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} {{qed}}
The group action axioms are investigated in turn. Let $\pi, \rho \in S_n$ and $n \in \N_n$. Thus: {{begin-eqn}} {{eqn | l = \pi * \paren {\rho * n}       | r = \pi * \map \rho n       | c = Definition of $*$ }} {{eqn | r = \map \pi {\map \rho n}       | c = Definition of $*$ }} {{eqn | r = \map {\paren {\pi \circ \rho} } n       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \paren {\pi \circ \rho} * n       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = I_{\N_n} * n       | r = \map {I_{\N_n} } n       | c = where $I_{\N_n}$ is the identity mapping on ${\N_n}$ }} {{eqn | r = n       | c = {{Defof|Identity Mapping}} }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. {{qed}}
By Group Action of Symmetric Group, $*: \map \Gamma S \times S \to S$ is indeed a group action Let $s, t \in S$. As $\map \Gamma S$ is the symmetric group on $S$, there exists a permutation $\pi \in \map \Gamma S$ such that: :$\map \pi s = t$ This holds for any $s, t \in S$. Thus: :$\forall t \in S: t \in \Orb s$ and so $S$ consists of a single orbit. Hence the result by definition of transitive group action. {{qed}}
The group action axioms are investigated in turn. Let $g_1, g_2 \in G$ and $s \in S$. Thus: {{begin-eqn}} {{eqn | l = g_1 * \paren {g_2 * s}       | r = g_1 * s       | c = Definition of $*$ }} {{eqn | r = s       | c = Definition of $*$ }} {{eqn | r = \paren {g_1 \circ g_2} * s       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = e * s       | r = s       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. {{qed}} Category:Group Actions eutj6bymo8jbd3w84zt2vao3iax559s
By definition: :$\operatorname{Orb} \left({s}\right) = \left\{{t \in S: \exists g \in G: g * s = t}\right\}$ By definition of the trivial group action: :$\forall g \in G: g * s = s$ Hence the result. {{qed}}
The group action axioms are investigated in turn. Let $g, h, a \in G$. Thus: {{begin-eqn}} {{eqn | l = g * \paren {h * a}       | r = g * \map {\rho_{h^{-1} } } a       | c = Definition of $*$ }} {{eqn | r = g * \paren {a \circ h^{-1} }       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = \map {\rho_{g^{-1} } } {a \circ h^{-1} }       | c = Definition of $*$ }} {{eqn | r = \paren {a \circ h^{-1} } \circ g^{-1}       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = a \circ \paren {h^{-1} \circ g^{-1} }       | c = {{GroupAxiom|1}} }} {{eqn | r = a \circ \paren {g \circ h}^{-1}       | c = Inverse of Group Product }} {{eqn | r = \map {\rho_{\paren {g \circ h}^{-1} } } a       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = \paren {g \circ h} * a       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = e * s       | r = \map {\rho_{e^{-1} } } s       | c = Definition of $*$ }} {{eqn | r = s \circ e^{-1}       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = s \circ e       | c = Inverse of Identity Element is Itself }} {{eqn | r = s       | c = {{GroupAxiom|2}} }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. {{qed}}
The group action axioms are investigated in turn. Let $g, h, a \in G$. Thus: {{begin-eqn}} {{eqn | l = g * \paren {h * a}       | r = g * \map {\lambda_h} a       | c = Definition of $*$ }} {{eqn | r = g * \paren {h \circ a}       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = \map {\lambda_g} {h \circ a}       | c = Definition of $*$ }} {{eqn | r = g \circ \paren {h \circ a}       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = \paren {g \circ h} \circ a       | c = Group Axiom $G1$: associativity }} {{eqn | r = \map {\lambda_{\paren {g \circ h} } } a       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = \paren {g \circ h} * a       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = e * s       | r = \map {\lambda_e} s       | c = Definition of $*$ }} {{eqn | r = e \circ s       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = s       | c = Group Axiom $G2$: properties of identity element }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. {{qed}}
Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | lo= \exists a \in G:       | l = h       | r = a \circ g^{-1}       | c = Group has Latin Square Property }} {{eqn | ll= \leadsto       | l = h       | r = \map {\rho_{g^{-1} } } a       | c = {{Defof|Right Regular Representation}} }} {{eqn | ll= \leadsto       | l = h       | r = g * a       | c = Definition of $*$ }} {{end-eqn}} $h$ is arbitrary, therefore the above holds for all $h \in G$. By definition of orbit: :$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$ That is: :$\Orb h = G$ Hence the result by definition of transitive group action. {{qed}}
Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | lo= \exists a \in G:       | l = h       | r = g \circ a       | c = Group has Latin Square Property }} {{eqn | ll= \leadsto       | l = h       | r = \map {\lambda_g} a       | c = {{Defof|Left Regular Representation}} }} {{eqn | ll= \leadsto       | l = h       | r = g * a       | c = Definition of $*$ }} {{end-eqn}} $h$ is arbitrary, therefore the above holds for all $h \in G$. By definition of orbit: :$\Orb h = \set {t \in G: \exists g \in G: g * h = t}$ That is: :$\Orb h = G$ Hence the result by definition of transitive group action. {{qed}}
Proof by Counterexample: For $G$ to be a transitive group action, the orbit of any element of $G$ needs to be the whole of $G$. Take $h = e$. Then: {{begin-eqn}} {{eqn | ll= \forall g \in G:       | l = g * e       | r = g \circ e \circ g^{-1}       | c =  }} {{eqn | r = g \circ g^{-1}       | c = {{GroupAxiom|2}} }} {{eqn | r = e       | c = {{GroupAxiom|3}} }} {{end-eqn}} Thus by definition of orbit: :$\Orb e = \set e$ Only when $G$ is the trivial group, that is: $G = \set e$, does $\Orb e = G$. Hence the result by definition of transitive group action. {{qed}}
For $G$ to be a trivial group action, the orbit of any element of $G$ is a singleton containing only that element. Take $h \in G$. Then: {{begin-eqn}} {{eqn | ll= \forall g \in G:       | l = g * h       | r = g \circ h \circ g^{-1}       | c =  }} {{eqn | r = h \circ g \circ g^{-1}       | c = {{Defof|Abelian Group}}: $g$ commutes with $h$ }} {{eqn | r = h \circ e       | c = Group axiom $G3$: properties of inverse element }} {{eqn | r = h       | c = Group axiom $G2$: properties of identity element }} {{end-eqn}} Thus by definition of orbit: :$\Orb h = \set h$ Hence the result by definition of trivial group action. {{qed}}
The group action axioms are investigated in turn. Let $h_1, h_2 \in H$ and $g \in G$. Thus: {{begin-eqn}} {{eqn | l = h_1 * \paren {h_2 * g}       | r = h_1 * \paren {\map {\lambda_{h_2} } g}       | c = Definition of $*$ }} {{eqn | r = h_1 * \paren {h_2 \circ g}       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = \map {\lambda_{h_1} } {h_2 \circ g}       | c = Definition of $*$ }} {{eqn | r = h_1 \circ \paren {h_2 \circ g}       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = \paren {h_1 \circ h_2} \circ g       | c = {{GroupAxiom|1}} }} {{eqn | r = \map {\lambda_{h_1 \circ h_2} } g       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = \paren {h_1 \circ h_2} * g       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = e * g       | r = \map {\lambda_e} g       | c = Definition of $*$ }} {{eqn | r = e \circ g       | c = {{Defof|Left Regular Representation}} }} {{eqn | r = g       | c = {{GroupAxiom|2}} }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. The group action axioms are thus seen to be fulfilled, and so $*$ is a group action. {{qed}}
The group action axioms are investigated in turn. Let $h_1, h_2 \in H$ and $g \in G$. Thus: {{begin-eqn}} {{eqn | l = h_1 * \paren {h_2 * g}       | r = h_1 * \paren {\map {\rho_{h_2^{-1} } } g}       | c = Definition of $*$ }} {{eqn | r = h_1 * \paren {g \circ h_2^{-1} }       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = \map {\rho_{h_1^{-1} } } {h_2^{-1} \circ g}       | c = Definition of $*$ }} {{eqn | r = \paren {g \circ h_2^{-1} } \circ h_1^{-1}       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = g \circ \paren {h_2^{-1} \circ h_1^{-1} }       | c = Group Axiom $G1$: Associativity }} {{eqn | r = g \circ \paren {h_1 \circ h_2}^{-1}       | c = Inverse of Group Product }} {{eqn | r = \map {\rho_{\paren {h_1 \circ h_2}^{-1} } } g       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = \paren {h_1 \circ h_2} * g       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = e * g       | r = \map {\rho_e} g       | c = Definition of $*$ }} {{eqn | r = g \circ e^{-1}       | c = {{Defof|Right Regular Representation}} }} {{eqn | r = g \circ e       | c = Inverse of Identity Element is Itself }} {{eqn | r = g       | c = Group Axiom $G2$: Properties of identity element }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. The group action axioms are thus seen to be fulfilled, and so $*$ is a group action. {{qed}}
From Group Action on Subgroup by Right Regular Representation it is established that $*$ is a group action. From Orbit of Group Action on Subgroup by Right Regular Representation is Right Coset: :$\forall x \in G: \Orb x = H x$ where $H x$ is the right coset of $H$ by $x$. From Right Coset Space forms Partition it is apparent that $H x \ne G$ unless $H = G$. Thus $\Orb x \ne G$ and the result follows by definition of transitive group action. {{qed}}
The inversion mapping on $G$ is the mapping $\iota: G \to G$ defined by: :$\forall g \in G: \map \iota g = g^{-1}$ where $g^{-1}$ is the inverse or $g$. By Inversion Mapping is Involution, $\iota$ is an involution: :$\forall g \in G: \map \iota {\map \iota g} = g$ The result follows from Involution is Permutation. {{qed}}
=== Proof of Surjection === Let $a \in G$. By definition of $\iota$: :$\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$ By Inverse of Inverse: :$\left({a^{-1}}\right)^{-1} = a$ Hence $a$ has a preimage. Since $a$ was arbitrary, $\iota$ is a surjection. === Proof of Injection === Suppose for some $a, b \in G$ that: :$\iota \left({a}\right) = \iota \left({b}\right)$ Then by the definition of $\iota$: :$a^{-1} = b^{-1}$ It follows from Inverse in Group is Unique that: :$a = b$ Hence $\iota$ is an injection. {{qed|lemma}} Hence by definition $\iota$ is a bijection. A bijection from a set to itself is by definition a permutation. {{qed}}
To show that $\iota$ is a mapping, it is sufficient to show that: :$\iota \left({a}\right) \ne \iota \left({b}\right) \implies a \ne b$: {{begin-eqn}} {{eqn | l = \iota \left({a}\right)       | o = \ne       | r = \iota \left({b}\right)       | c = by hypothesis }} {{eqn | ll= \implies       | l = a^{-1}       | o = \ne       | r = b^{-1}       | c = by definition of $\iota$ }} {{eqn | ll= \implies       | l = a \circ a^{-1} \circ b       | o = \ne       | r = a \circ b^{-1} \circ b       | c = Cancellation Laws and Group Axiom $G1$: Associativity }} {{eqn | ll= \implies       | l = e \circ b       | o = \ne       | r = a \circ e       | c = Group Axiom $G3$: Properties of Inverse Element }} {{eqn | ll= \implies       | l = b       | o = \ne       | r = a       | c = Group Axiom $G2$: Properties of Identity Element }} {{end-eqn}} {{qed}} Category:Inversion Mappings joe5rd10ivc7k8jnxmv77ctakpp05ft
The group action axioms are investigated in turn. Let $g, h \in G$ and $s, t \in S$. Thus: {{begin-eqn}} {{eqn | l = g * \tuple {h * \tuple {s, t} }       | r = g * \tuple {h *_S s, h *_T t}       | c = Definition of $*$ }} {{eqn | r = \tuple {g *_S \tuple {h *_S s}, g *_T \tuple {h *_T t} }       | c = Definition of $*$ }} {{eqn | r = \tuple {\paren {g \circ h} *_S s, \paren {g \circ h} *_T t}       | c = {{GroupActionAxiom|1}} for both $*_S$ and $*_T$ }} {{eqn | r = \paren {g \circ h} * \tuple {s, t}       | c = Definition of $*$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|1}} holds. Then: {{begin-eqn}} {{eqn | l = e * \tuple {s, t}       | r = \tuple {e *_S s, e *_T t}       | c = Definition of $*$ }} {{eqn | r = \tuple {s, t}       | c = {{GroupActionAxiom|2}} for both $*_S$ and $*_T$ }} {{end-eqn}} demonstrating that {{GroupActionAxiom|2}} holds. The group action axioms are thus seen to be fulfilled, and so $*$ is a group action. {{qed}}
By definition, the stabilizer of an element $x$ of $S$ is defined as: :$\Stab x := \set {g \in G: g * x = x}$ where $*$ denotes the group action. So: {{begin-eqn}} {{eqn | l = \Stab {s, t}       | r = \set {g \in G: g * \tuple {s, t} = \tuple {s, t} }       | c = {{Defof|Stabilizer}} }} {{eqn | r = \set {g \in G: \tuple {g *_S s, g *_T t} = \tuple {s, t} }       | c = Definition of $*$ }} {{eqn | r = \set {g \in G: g *_S s = s \land g *_T t = t}       | c = {{Defof|Cartesian Product}} }} {{eqn | r = \set {g \in G: g \in \Stab s \land g \in \Stab t}       | c = {{Defof|Stabilizer}} }} {{eqn | r = \Stab s \cap \Stab t       | c = {{Defof|Set Intersection}} }} {{end-eqn}} {{qed}}
We list out all the left cosets of $H \cap K$ in $H$: :$H / \paren {H \cap K} = \set {h_n \paren {H \cap K}: h_n \in H, n \in I}$ where $I$ is some finite indexing set. For each pair $h_i, h_j \in H \subseteq G$, where $i \ne j$: :$h_i^{-1} h_j \notin H \cap K \quad$ Cosets are Equal iff Product with Inverse in Subgroup Since $H$ is a subgroup, $h_i^{-1} h_j \in H$. Thus: :$h_i^{-1} h_j \notin K$ By Cosets are Equal iff Product with Inverse in Subgroup, $h_i K$ and $h_j K$ are different left cosets. Hence there must be at least $\index H {H \cap K}$ left cosets of $K$ in $G$, namely $h_n K$ for each $n \in I$. It follows that $\index H {H \cap K} \le \index G K$. {{qed|lemma}} === Equality === Suppose equality holds. Then the construction above gives all left cosets of $K$ in $G$. For any $x \in G$, $x \in h_i K$ for some $h_i \in H$. Hence $G \subseteq H K$. The inclusion $H K \subseteq G$ is seen from {{GroupAxiom|0}}. Hence $G = H K$ by definition of set equality. {{qed|lemma}} Now suppose $G = H K$. To show equality we must show that the construction above gives all left cosets of $K$ in $G$. Let $g K$ be a left coset of $K$ in $G$, where $x \in G$. Since $G = H K$: :$\exists h \in H, k \in K: g = h k$ Then by Left Coset Equals Subgroup iff Element in Subgroup: :$g K = h \paren {k K} = h K$ Since $h \in H$: :$\exists i \in I: h \in h_i \paren {H \cap K}$ By Element in Left Coset iff Product with Inverse in Subgroup: :$h^{-1} h_i \in H \cap K \subseteq K$ Thus by Left Cosets are Equal iff Product with Inverse in Subgroup: :$h_i K = h K = g K$ Hence all left cosets of $K$ in $G$ are obtained in the construction. This gives equality. {{qed}}
From List of Elements in Finite Cyclic Group, the elements of $G$ are: :$\set {g^k: g \in G, 0 \le k < n}$ From Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order, $g^k$ generates $G$ {{iff}} $k \perp n$. The result follows by definition of the Euler $\phi$ function. {{qed}}
{{BeginTableau|p \lor q, \neg p \vdash q}} {{Premise|1|p \lor q}} {{Premise|2|\neg p}} {{Assumption|3|p}} {{SequentIntro|4|2|p \implies q|2|False Statement implies Every Statement}} {{ModusPonens|5|2, 3|q|4|3}} {{Assumption|6|q}} {{ProofByCases|7|1, 2|q|1|3|5|6|6}} {{EndTableau}} {{qed}}
We apply the Method of Truth Tables to the proposition. $\begin{array}{|ccc|cc||c|} \hline p & \lor & q & \neg & p & q\\ \hline F & F & F & T & F & F \\ F & T & T & T & F & T \\ T & T & F & F & T & F \\ T & T & T & F & T & T \\ \hline \end{array}$ As can be seen, when $p \lor q$ is true, and so is $\neg p$, then $q$ is also true. {{qed}}
=== Sufficient Condition === Let $\exists x: \map S x$. Suppose $\mathbf A$ and $\mathbf E$ are both true. As $\mathbf A$ is true, then by Modus Ponendo Ponens: :$\map P x$ As $\mathbf E$ is true, then by Modus Ponendo Ponens: :$\neg \map P x$ It follows by Proof by Contradiction that $\mathbf A$ and $\mathbf E$ are not both true. Thus, by definition, $\mathbf A$ and $\mathbf E$ are contrary statements. {{qed|lemma}} === Necessary Condition === Let $\mathbf A$ and $\mathbf E$ be contrary statements. Suppose: :$\neg \exists x: \map S x$ that is, $\map S x$ is vacuous. From Denial of Existence: :$\forall x: \neg \map S x \dashv \vdash \neg \exists x: \map S x$ it follows that $\forall x: \map S x$ is false. From False Statement implies Every Statement: :$\forall x: \map S x \implies \map P x$ is true, and: :$\forall x: \map S x \implies \neg \map P x$ is also true. Thus, by definition, $\mathbf A$ and $\mathbf E$ are not contrary statements. It follows by Proof by Contradiction that $\exists x: \map S x$. {{qed}}
=== Sufficient Condition === Let $\exists x: \map S x$. Suppose $\mathbf I$ and $\mathbf O$ are both false. As $\mathbf I$ is false, then by the Rule of Conjunction: :$\neg \map P x$ As $\mathbf O$ is false, then by the Rule of Conjunction: :$\neg \neg \map P x$ and so by Double Negation: :$\map P x$ It follows by Proof by Contradiction that $\mathbf I$ and $\mathbf O$ are not both false. Thus, by definition, $\mathbf I$ and $\mathbf O$ are subcontrary statements. {{qed|lemma}} === Necessary Condition === Let $\mathbf I$ and $\mathbf O$ be subcontrary statements Suppose $\neg \exists x: \map S x$. From the definition of logical conjunction, it follows that: :$\neg \paren {\exists x: \map S x \land \map P x}$ Similarly from the definition of logical conjunction, it follows that: :$\neg \paren {\exists x: \map S x \land \neg \map P x}$ That is, both $\mathbf I$ and $\mathbf O$ are false. So, by definition, $\mathbf I$ and $\mathbf O$ are not subcontrary. It follows by Proof by Contradiction that $\exists x: \map S x$. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \mathbf A       | c =  }} {{eqn | ll= \leadsto       | l = \forall x:       | o =        | r = \map S x \implies \map P x       | c = Definition of $\mathbf A$ }} {{eqn | ll= \leadsto       | l = \forall x:       | o =        | r = \neg \paren {\map S x \land \neg \map P x}       | c = Implication Equivalent to Negation of Conjunction with Negative }} {{eqn | ll= \leadsto       | l = \neg \exists x:       | o =        | r = \map S x \land \neg \map P x       | c = De Morgan's Laws: Denial of Existence }} {{eqn | ll= \leadsto       | o =        | r = \neg \mathbf O       | c = Definition of $\mathbf O$ }} {{end-eqn}} The argument reverses: {{begin-eqn}} {{eqn | o =        | r = \mathbf O       | c =  }} {{eqn | ll= \leadsto       | l = \exists x:       | o =        | r = \map S x \land \neg \map P x       | c = Definition of $\mathbf O$ }} {{eqn | ll= \leadsto       | l = \exists x:       | o =        | r = \neg \paren {\map S x \implies \map P x}       | c = Conjunction with Negative Equivalent to Negation of Implication }} {{eqn | ll= \leadsto       | l = \neg \forall x:       | o =        | r = \map S x \implies \map P x       | c = De Morgan's Laws: Denial of Universality }} {{eqn | ll= \leadsto       | o =        | r = \neg \mathbf A       | c = Definition of $\mathbf A$ }} {{end-eqn}} The result follows by definition of contradictory. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \mathbf E       | c =  }} {{eqn | ll= \therefore       | l = \forall x:       | o =        | r = \map S x \implies \neg \map P x       | c = Definition of $\mathbf E$ }} {{eqn | ll= \therefore       | l = \forall x:       | o =        | r = \neg \paren {\map S x \land \map P x}       | c = Modus Ponendo Tollens }} {{eqn | ll= \therefore       | l = \neg \exists x:       | o =        | r = \map S x \land \map P x       | c = De Morgan's Laws: Denial of Existence }} {{eqn | ll= \therefore       | o =        | r = \neg \mathbf I       | c = Definition of $\mathbf I$ }} {{end-eqn}} The argument reverses: {{begin-eqn}} {{eqn | o =        | r = \mathbf I       | c =  }} {{eqn | ll= \therefore       | l = \exists x:       | o =        | r = \map S x \land \map P x       | c = Definition of $\mathbf I$ }} {{eqn | ll= \therefore       | l = \exists x:       | o =        | r = \neg \paren {\map S x \implies \neg \map P x}       | c = Conjunction Equivalent to Negation of Implication of Negative }} {{eqn | ll= \therefore       | l = \neg \forall x:       | o =        | r = \map S x \implies \neg \map P x       | c = De Morgan's Laws: Denial of Universality }} {{eqn | ll= \therefore       | o =        | r = \neg \mathbf E       | c = Definition of $\mathbf E$ }} {{end-eqn}} The result follows by definition of contradictory. {{qed}}