hcju/proofwikiqa huggingface dataset:
=== Sufficient Condition === Let $\exists x: \map S x$. Let $\map {\mathbf A} {S, P}$ be true. As $\map {\mathbf A} {S, P}$ is true, then by Modus Ponendo Ponens: :$\map P x$ From the Rule of Conjunction: :$\map S x \land \map P x$ Thus $\map {\mathbf I} {S, P}$ holds. So by the Rule of Implication: :$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$ {{qed|lemma}} === Necessary Condition === Let $\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$. Suppose: :$\neg \exists x: \map S x$ that is, $\map S x$ is vacuous. From De Morgan's Laws: 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. So $\map {\mathbf A} {S, P}$ holds. Again, $\neg \exists x: \map S x$. Then by the Rule of Conjunction: :$\neg \paren {\exists x: \map S x \land \map P x}$ That is, $\mathbf I$ does not hold. So $\map {\mathbf A} {S, P}$ is true and $\map {\mathbf I} {S, P}$ is false. This contradicts $\map {\mathbf A} {S, P} \implies \map {\mathbf I} {S, P}$ by definition of implication. Thus $\exists x: \map S x$ must hold. {{qed}}
=== Sufficient Condition === Let $\exists x: \map S x$. Let $\map {\mathbf E} {S, P}$ be true. As $\map {\mathbf E} {S, P}$ is true, then by Modus Ponendo Ponens: :$\neg \map P x$ From the Rule of Conjunction: :$\map S x \land \neg \map P x$ Thus $\map {\mathbf O} {S, P}$ holds. So by the Rule of Implication: :$\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$ {{qed|lemma}} === Necessary Condition === Let $\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$. {{AimForCont}}: :$\neg \exists x: \map S x$ that is, $\map S x$ is vacuous. From De Morgan's Laws: 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: :$\forall x: \map S x \implies \neg \map P x$ is true. So $\map {\mathbf E} {S, P}$ holds. Again, $\neg \exists x: \map S x$. Then by the Rule of Conjunction: :$\neg \paren {\exists x: \map S x \land \neg \map P x}$ That is, $\map {\mathbf O} {S, P}$ does not hold. So $\map {\mathbf E} {S, P}$ is true and $\map {\mathbf O} {S, P}$ is false. This contradicts $\map {\mathbf E} {S, P} \implies \map {\mathbf O} {S, P}$ by definition of implication. Thus $\exists x: \map S x$ must hold. {{qed}}
Let $x$ be an object variable from the universe of '''rational beings'''. Let $\map H x$ denote the propositional function ''$x$ is '''human'''''. Let $\map M x$ denote the propositional function ''$x$ is '''mortal'''''. Let $S$ be a proper name that denotes {{AuthorRef|Socrates}}. The argument can then be expressed as: {{begin-eqn}} {{eqn | n = 1       | lo= \forall x:       | l = \map H x       | o = \implies       | r = \map M x       | c =  }} {{eqn | ll= \therefore       | l = \map H S       | o = \implies       | r = \map M S       | c = Universal Instantiation }} {{eqn | n = 2       | l = \map H S       | o =        | c =  }} {{eqn | n = 3       | ll= \therefore       | l = \map M S       | o =        | c = Modus Ponendo Ponens }} {{end-eqn}} That is:  :''{{AuthorRef|Socrates}} is mortal.'' {{qed}}
By hypothesis, $f'_+ \left({a}\right)$ exists. First we note that $a$ cannot be the right hand end point of $I$ because values in $I$ greater than $a$ need to exist for $f'_+ \left({a}\right)$ to exist. We form the following expression: :$\displaystyle \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)$ We need to show that it is defined and to find its value. We find: {{begin-eqn}} {{eqn | o =        | r = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)       | c =  }} {{eqn | r = \lim_{x \mathop \to a^+} \left({\frac {f \left({x}\right) − f \left({a}\right)} {x − a} \left({x − a}\right)}\right)       | c = where the denominator is unequal to $0$ since $x > a$ }} {{eqn | r = \lim_{x \mathop \to a^+} \left({\frac {f \left({x}\right) − f \left({a}\right)} {x − a} }\right) \lim_{x \mathop \to a^+} \left({x−a}\right)       | c = Product Rule for Limits of Functions since (see the next step) the two limits exist }} {{eqn | r = f'_+ \left({a}\right) \times 0       | c = {{Defof|Right-Hand Derivative}} }} {{eqn | r = 0 }} {{end-eqn}} Note that this proves that $\displaystyle \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)$ exists. We continue by manipulating the result above: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right)       | r = 0 }} {{eqn | ll= \iff       | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right) + f \left({a}\right) - f \left({a}\right)       | r = 0 }} {{eqn | ll= \iff       | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right)}\right) + \lim_{x \mathop \to a^+} f \left({a}\right) - f \left({a}\right)       | r = 0 }} {{eqn | ll= \iff       | l = \lim_{x \mathop \to a^+} \left({f \left({x}\right) − f \left({a}\right) + f \left({a}\right)}\right) - f \left({a}\right)       | r = 0       | c = Sum Rule for Limits of Functions since the two limits in the previous expression exist }} {{eqn | ll= \iff       | l = \lim_{x \mathop \to a^+} f \left({x}\right) - f \left({a}\right)       | r = 0 }} {{eqn | ll= \iff       | l = \lim_{x \mathop \to a^+} f \left({x}\right)       | r = f \left({a}\right) }} {{end-eqn}} 	 which means that $f$ is right-continuous at $a$. {{qed}} Category:Continuous Functions 9no786rkltfcbpsn4curh9qhem8d0wf
By hypothesis, $\map {f'_-} a$ exists. First we note that $a$ cannot be the left-hand end point of $I$ because values in $I$ less than $a$ need to exist for $\map {f'_-} a$ to exist. We form the following expression: :$\displaystyle \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}$ We need to show that it is defined and to find its value. We find: {{begin-eqn}} {{eqn | o =        | r = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}       | c =  }} {{eqn | r = \lim_{x \mathop \to a^-} \paren {\frac {\map f x − \map f a} {x − a} \paren {x − a} }       | c = where the denominator is unequal to $0$ since $x < a$ }} {{eqn | r = \lim_{x \mathop \to a^-} \paren {\frac {\map f x − \map f a} {x − a} } \lim_{x \mathop \to a^-} \paren {x − a}       | c = Product Rule for Limits of Functions since (see the next step) the two limits exist }} {{eqn | r = \map {f'_-} a \times 0       | c = {{Defof|Left-Hand Derivative}} }} {{eqn | r = 0 }} {{end-eqn}} Note that this proves that $\displaystyle \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}$ exists. We continue by manipulating the result above: {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a}       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a} + \map f a - \map f a       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a} + \lim_{x \mathop \to a^-} \map f a - \map f a       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \lim_{x \mathop \to a^-} \paren {\map f x − \map f a + \map f a} - \map f a       | r = 0       | c = Sum Rule for Limits of Functions since the two limits in the previous expression exist }} {{eqn | ll= \leadstoandfrom       | l = \lim_{x \mathop \to a^-} \map f x - \map f a       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \lim_{x \mathop \to a^-} \map f x       | r = \map f a }} {{end-eqn}} 	 which means that $f$ is left-continuous at $a$. {{qed}} Category:Continuous Functions a5ozw10iwkagg95a83w0ncvfkxawznn
By Left-Hand Differentiable Function is Left-Continuous, $f$ is left-continuous at $a$. By Right-Hand Differentiable Function is Right-Continuous, $f$ is right-continuous at $a$. By Limit iff Limits from Left and Right, $f$ is continuous at $a$ since $f$ is left- and right-continuous at $a$. {{qed}} Category:Continuous Functions jlvd6r0vqqj5okpdfr9jep3h7cqz3k6
=== Necessary Condition === Let $\mathbf A \left({S, P}\right)$ and $\mathbf E \left({S, P}\right)$ both be false. {{begin-eqn}} {{eqn | n = 1       | o =       | r = \neg \mathbf A \left({S, P}\right) \land \neg \mathbf E \left({S, P}\right) }} {{eqn | n = 2       | o = \leadsto       | r = \neg \mathbf A \left({S, P}\right)       | c = Rule of Simplification: from $(1)$ }} {{eqn | n = 3       | o = \leadsto       | r = \mathbf O \left({S, P}\right)       | c = Universal Affirmative and Particular Negative are Contradictory: from $(2)$ }} {{eqn | n = 4       | o = \leadsto       | r = \neg \mathbf E \left({S, P}\right)       | c = Rule of Simplification: from $(1)$ }} {{eqn | n = 5       | o = \leadsto       | r = \mathbf I \left({S, P}\right)       | c = Particular Affirmative and Universal Negative are Contradictory: from $(4)$ }} {{eqn | n = 6       | o = \leadsto       | r = \mathbf I \left({S, P}\right) \land \mathbf O \left({S, P}\right)       | c = Rule of Conjunction: from $(5)$ and $(3)$ }} {{end-eqn}} Hence $\mathbf I \left({S, P}\right)$ and $\mathbf O \left({S, P}\right)$ are both true. {{qed|lemma}} === Sufficient Condition === Let $\mathbf I \left({S, P}\right)$ and $\mathbf O \left({S, P}\right)$ both be true. {{begin-eqn}} {{eqn | n = 1       | o =       | r = \mathbf I \left({S, P}\right) \land \mathbf O \left({S, P}\right) }} {{eqn | n = 2       | o = \leadsto       | r = \mathbf I \left({S, P}\right)       | c = Rule of Simplification: from $(1)$ }} {{eqn | n = 3       | o = \leadsto       | r = \neg \mathbf E \left({S, P}\right)       | c = Particular Affirmative and Universal Negative are Contradictory: from $(2)$ }} {{eqn | n = 4       | o = \leadsto       | r = \mathbf O \left({S, P}\right)       | c = Rule of Simplification: from $(1)$ }} {{eqn | n = 5       | o = \leadsto       | r = \neg \mathbf A \left({S, P}\right)       | c = Universal Affirmative and Particular Negative are Contradictory: from $(4)$ }} {{eqn | n = 6       | o = \leadsto       | r = \neg \mathbf A \left({S, P}\right) \land \neg \mathbf E \left({S, P}\right)       | c = Rule of Conjunction: from $(5)$ and $(3)$ }} {{end-eqn}} Hence $\mathbf A \left({S, P}\right)$ and $\mathbf E \left({S, P}\right)$ are both false. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \map {\mathbf I} {S, P}       | c =  }} {{eqn | ll= \leadsto       | o =        | r = \exists x: \map S x \land \map P x       | c = {{Defof|Particular Affirmative}} }} {{eqn | ll= \leadsto       | o =        | r = \exists x: \map P x \land \map S x       | c = Conjunction is Commutative }} {{eqn | ll= \leadsto       | o =        | r = \map {\mathbf I} {P, S}       | c = {{Defof|Particular Affirmative}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \mathbf E \left({S, P}\right)       | c =  }} {{eqn | o = \implies       | r = \forall x: S \left({x}\right) \implies \neg P \left({x}\right)       | c = Definition of Universal Negative }} {{eqn | o = \implies       | r = \forall x: \neg \left({S \left({x}\right) \land P \left({x}\right)}\right)       | c = Modus Ponendo Tollens }} {{eqn | o = \implies       | r = \forall x: \neg \left({P \left({x}\right) \land S \left({x}\right)}\right)       | c = Conjunction is Commutative }} {{eqn | o = \implies       | r = \forall x: P \left({x}\right) \implies \neg S \left({x}\right)       | c = Modus Ponendo Tollens }} {{eqn | o = \implies       | r = \mathbf E \left({P, S}\right)       | c = Definition of Universal Negative }} {{end-eqn}} {{qed}}
From Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous: :$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$ From Law of Simple Conversion of I: :$\paren {\exists x: \map S x \land \map P x} \implies \paren {\exists x: \map P x \land \map S x}$ Hence the result. {{qed}}
Recall the four figures of the categorical syllogism: :$\begin{array}{r|rl} \text I & & \\ \hline \\ \text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\ \text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\ \hline \\ \text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\ \end{array} \qquad \begin{array}{r|rl} \text {II} & & \\ \hline \\ \text{Major Premise}: & \mathbf \Phi_1 & \tuple {P, M} \\ \text{Minor Premise}: & \mathbf \Phi_2 & \tuple {S, M} \\ \hline \\ \text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\ \end{array}$ :$\begin{array}{r|rl} \text {III} & & \\ \hline \\ \text{Major Premise}: & \mathbf \Phi_1 & \tuple {M, P} \\ \text{Minor Premise}: & \mathbf \Phi_2 & \tuple {M, S} \\ \hline \\ \text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\ \end{array} \qquad \begin{array}{r|rl} \text {IV} & & \\ \hline \\ \text{Major Premise}: & \mathbf \Phi_1 & \tuple {P, M} \\ \text{Minor Premise}: & \mathbf \Phi_2 & \tuple {M, S} \\ \hline \\ \text{Conclusion}: & \mathbf \Phi_3 & \tuple {S, P} \\ \end{array}$ By the definition of a categorical syllogism, the following are fixed: :$(1): \quad$ The order of the categorical statements: major premise, then minor premise, then conclusion. :$(2): \quad$ The structure of the conclusion: the secondary term then the primary term. :$(3): \quad$ The content of the two premises: the major premise contains the primary term and the middle term, while the minor premise contains the middle term and the secondary term. The things that can be varied are: :$(A): \quad$ The specific nature of each of the categorical statements: each can be any of four types: $\mathbf A$, $\mathbf E$, $\mathbf I$ or $\mathbf O$. :$(B): \quad$ Within the major and minor premises, the order of the two terms: the middle term can come either first or second. The order of the terms within the premises determines the figure of the categorical syllogism. For each of the two premises there are two options for the position of the middle term. By the Product Rule for Counting it follows that there are $2 \times 2 = 4$ figures. In each figure, there are $3$ categorical statements. Each can be of any of $4$ types. Hence by the Product Rule for Counting, there are $4 \times 4 \times 4 = 64$ possible standard instances for each figure. For all $4$ figures, by the Product Rule for Counting, it follows that there are $4 \times 64 = 256$ standard instances in total. {{qed}}
From Elimination of all but 24 Categorical Syllogisms as Invalid, all but these $24$ patterns have been shown to be invalid. It remains to be shown that these remaining syllogisms are in fact valid. {{ProofWanted|Considerable work to be done yet.}}
Suppose both premises of $Q$ are particular. Then the pattern of $Q$ is one of $\text{II}x$, $\text{IO}x$, $\text{OI}x$ or $\text{OO}x$, where $x$ is the conclusion. $\text{I}$ is neither universal nor negative. Thus the $\text{II}x$ pattern does not distribute the middle term of $Q$. So $\text{II}x$ violates the rule Middle Term of Valid Categorical Syllogism is Distributed at least Once. $\text{O}$ is negative. Thus $\text{OO}x$ violates the rule No Valid Categorical Syllogism contains two Negative Premises. It remains to investigate $\text{IO}x$ and $\text{OI}x$. By Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative, the conclusion of $Q$ is negative, either $\text{E}$ or $\text{O}$. By the definition of Distributed Predicate of Categorical Syllogism, the predicate of the conclusion of $Q$ is distributed. Thus, by construction, the primary term $P$ of $Q$ is distributed. By Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise, it follows that $P$ is distributed in the major premise of $Q$. This eliminates $\text{IO}x$ as the particular affirmative $\text{I}$ distributes neither its subject nor its predicate. The final pattern to be investigated is $\text{OI}x$. By definition, $\text{O}$ distributes only its predicate. As $\text{O}$ needs to distribute the primary term $P$, it cannot also distribute the middle term $M$ of $Q$. But the particular affirmative $\text{I}$ (as has been seen above) also does not distribute $M$. Thus $M$ remains undistributed, so violating the rule Middle Term of Valid Categorical Syllogism is Distributed at least Once. This eliminates $\text{OI}x$. {{qed}}
Let the major premise of $Q$ be denoted $\text{Maj}$. Let the minor premise of $Q$ be denoted $\text{Min}$. Let the conclusion of $Q$ be denoted $\text{C}$. From No Valid Categorical Syllogism contains two Particular Premises, either $\text{Maj}$ or $\text{Min}$ has to be universal. Let the other premise of $Q$ be particular. Suppose $\text{C}$ is the universal affirmative $\mathbf A \left({S, P}\right)$. From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative it follows that $\text{Maj}$ and $\text{Min}$ are also both affirmative. The only patterns fitting the criteria are $\text{AI}x$ and $\text{IA}x$, both of which distribute only one term in $Q$. We have that $\text{C}$ distributes $S$. From Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise, $S$ must also be distributed in $\text{Min}$. From Middle Term of Valid Categorical Syllogism is Distributed at least Once, the middle term $M$ of $Q$ needs to be distributed. Thus we require two distributed terms in the premises for $\text{C}$ to be the universal affirmative. So if $\text{C}$ is affirmative, neither premise of $Q$ can be particular. From this contradiction it follows that $\text{C}$ is not $\mathbf A \left({S, P}\right)$. Suppose $\text{C}$ is the universal negative $\mathbf E \left({S, P}\right)$. In this case, both $S$ and $P$ are distributed in $C$, as the universal negative distributes both $S$ and $P$. 	 By Middle Term of Valid Categorical Syllogism is Distributed at least Once, the middle term $M$ of $Q$ needs to be distributed. 	 Thus there are three terms needing to be distributed in two premises. So at least one premises needs to be the universal negative. The other premise is not $\mathbf O$ as this violates No Valid Categorical Syllogism contains two Negative Premises. But it also is not $\mathbf I$, as $\mathbf I$ does not distribute any terms. So if $\text{C}$ is the universal negative, neither premise can be particular. From this contradiction it follows that $\text{C}$ is not $\mathbf E \left({S, P}\right)$. {{qed}}
There are $64$ patterns of categorical syllogism per figure: :$\begin{array}{cccc} AAA & AAE & AAI & AAO \\ AEA & AEE & AEI & AEO \\ AIA & AIE & AII & AIO \\ AOA & AOE & AOI & AOO \\ \end{array} \qquad \begin{array}{cccc} EAA & EAE & EAI & EAO \\ EEA & EEE & EEI & EEO \\ EIA & EIE & EII & EIO \\ EOA & EOE & EOI & EOO \\ \end{array}$ :$\begin{array}{cccc} IAA & IAE & IAI & IAO \\ IEA & IEE & IEI & IEO \\ IIA & IIE & III & IIO \\ IOA & IOE & IOI & IOO \\ \end{array} \qquad \begin{array}{cccc} OAA & OAE & OAI & OAO \\ OEA & OEE & OEI & OEO \\ OIA & OIE & OII & OIO \\ OOA & OOE & OOI & OOO \\ \end{array}$ From No Valid Categorical Syllogism contains two Negative Premises, all those whose patterns start $EE$, $EO$, $OE$ and $OO$ can be eliminated: :$\begin{array}{cccc} AAA & AAE & AAI & AAO \\ AEA & AEE & AEI & AEO \\ AIA & AIE & AII & AIO \\ AOA & AOE & AOI & AOO \\ \end{array} \qquad \begin{array}{cccc} EAA & EAE & EAI & EAO \\   &  &  &  \\ EIA & EIE & EII & EIO \\  &  &   &   \\ \end{array}$ :$\begin{array}{cccc} IAA & IAE & IAI & IAO \\ IEA & IEE & IEI & IEO \\ IIA & IIE & III & IIO \\ IOA & IOE & IOI & IOO \\ \end{array} \qquad \begin{array}{cccc} OAA & OAE & OAI & OAO \\   &   &   &   \\ OIA & OIE & OII & OIO \\   &   &  &   \\ \end{array}$ From No Valid Categorical Syllogism contains two Particular Premises, all those whose patterns start $II$, $IO$ and $OI$ and $OO$ can be eliminated: :$\begin{array}{cccc} AAA & AAE & AAI & AAO \\ AEA & AEE & AEI & AEO \\ AIA & AIE & AII & AIO \\ AOA & AOE & AOI & AOO \\ \end{array} \qquad \begin{array}{cccc} EAA & EAE & EAI & EAO \\   &  &  &  \\ EIA & EIE & EII & EIO \\ \end{array}$ :$\begin{array}{cccc} IAA & IAE & IAI & IAO \\ IEA & IEE & IEI & IEO \\ \end{array} \qquad \begin{array}{cccc} OAA & OAE & OAI & OAO \\ \end{array}$ From Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative, further patterns can be eliminated: :$\begin{array}{cccc} AAA & & AAI & \\  & AEE & & AEO \\ AIA & & AII & \\  & AOE & & AOO \\ \end{array} \qquad \begin{array}{cccc}  & EAE & & EAO \\  &  &  &  \\  & EIE & & EIO \\ \end{array}$ :$\begin{array}{cccc} IAA & & IAI & \\  & IEE & & IEO \\ \end{array} \qquad \begin{array}{cccc}  & OAE & & OAO \\ \end{array}$ From No Valid Categorical Syllogism with Particular Premise has Universal Conclusion, all those whose patterns match that condition can be eliminated: :$\begin{array}{cccc} AAA & & AAI & \\  & AEE & & AEO \\  & & AII & \\  & & & AOO \\ \end{array} \qquad \begin{array}{cccc}  & EAE & & EAO \\  &  &  &  \\  & & & EIO \\ \end{array}$ :$\begin{array}{cccc}  & & IAI & \\  & & & IEO \\ \end{array} \qquad \begin{array}{cccc}  & & & OAO \\ \end{array}$ Thus there are $12$ patterns remaining. Each one may apply to any one of the $4$ figures Thus there are no more than $48$ valid patterns of categorical syllogism. {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations. :$\begin{array}{|ccc|c|ccc|} \hline (p & \land & q) & \implies & (p & \lor & q) \\ \hline F & F & F & T & F & F & F \\ F & F & T & T & F & T & T \\ T & F & F & T & T & T & F \\ T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
:400px As shown above: :Let the angle between $\mathbf u$ and $\mathbf v$ be $\gamma$. :Let the angle between $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ be $\alpha$. :Let the angle between $\mathbf u$ and $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ be $\beta$. Note that $\left\Vert{ \mathbf u }\right\Vert \mathbf v$ is $\mathbf v$ multiplied by the length of $\mathbf u$. By Vector Times Magnitude Same Length As Magnitude Times Vector the vectors $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ have equal length. So $\left\Vert{\mathbf u}\right\Vert \mathbf v$, $\left\Vert{\mathbf v}\right\Vert \mathbf u$ and $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ make an isosceles triangle. Therefore: {{begin-eqn}} {{eqn | l = 2 \beta + \alpha       | r = 180^\circ }} {{eqn | l = \beta       | r = 90^\circ - \frac 1 2 \alpha }} {{eqn | l = 2 \beta       | r = 180^\circ - \alpha }} {{end-eqn}} But since $\mathbf v$ and $\left\Vert{ \mathbf u }\right\Vert \mathbf v$ are parallel, we also have: {{begin-eqn}} {{eqn | l = \delta       | r = \alpha       | c = Parallelism implies Equal Corresponding Angles }} {{eqn | l = \delta + \gamma       | r = 180^\circ }} {{eqn | l = \alpha + \gamma       | r = 180^\circ }} {{eqn | l = \gamma       | r = 180^\circ - \alpha }} {{end-eqn}} Thus $\gamma = 2 \beta$, and the result follows. {{qed}}
The vectors $\left\Vert{\mathbf u}\right\Vert \mathbf v$ and $\left\Vert{\mathbf v}\right\Vert \mathbf u$ have equal length from Vector Times Magnitude Same Length As Magnitude Times Vector. Thus $\left\Vert{\mathbf u}\right\Vert \mathbf v + \left\Vert{\mathbf v}\right\Vert \mathbf u$ is the diagonal of a rhombus.   The result follows from Diagonals of Rhombus Bisect Angles. {{qed}}