hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \cos^n \theta       | r = \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}^n       | c = De Moivre's Theorem }} {{eqn | r = \frac {\paren {e^{i \theta} + e^{-i \theta} }^n} {2^n}       | c =  }} {{eqn | r = \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - k} i \theta} e^{-k i \theta}       | c = Binomial Theorem }} {{eqn | r = \frac 1 {2^n} \sum_{k \mathop = 0}^n \binom n k e^{\paren {n - 2 k} i \theta}       | c =  }} {{end-eqn}} Matching up terms from the beginning of this expansion with those from the end: {{begin-eqn}} {{eqn | l = 2^n \cos^n \theta       | r = e^{n i \theta} + \binom n 1 e^{\paren {n - 2} i \theta} + \binom n 2 e^{\paren {n - 4} i \theta} + \cdots       | c =  }} {{eqn | o =       | ro= +       | r = \binom n {n - 2} e^{-\paren {n - 4} i \theta} + \binom n {n - 1} e^{-\paren {n - 2} i \theta} + e^{-n i \theta}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {2^n} 2 \cos^n \theta       | r = \paren {\frac {e^{n i \theta} + e^{-n i \theta} } 2} + \binom n 1 \paren {\frac {e^{\paren {n - 2} i \theta} + e^{-\paren {n - 2} i \theta} } 2}       | c =  }} {{eqn | o =       | ro= +       | r = \binom n 2 \paren {\frac {e^{\paren {n - 4} i \theta} + e^{-\paren {n - 4} i \theta} } 2} + \cdots       | c =  }} {{end-eqn}} Thus: :$\cos^n \theta = \dfrac 1 {2^{n - 1} } \paren {\cos n \theta + n \cos \paren {n - 2} \theta + \dfrac {n \paren {n - 1} } {2!} \cos \paren {n - 4} \theta + \cdots + R_n}$ Now to determine $R_n$. The middle term of the sequence $0, 1, \ldots, n$ is $\dfrac n 2$. So when $k = \dfrac n 2$ we have $n - 2k = 0$ and $n - k = n - \dfrac n 2 = \dfrac n 2$. Thus: {{begin-eqn}} {{eqn | l = \binom n k       | r = \frac {n!} {\paren {\paren {n / 2}!} \paren {\paren {n - n / 2}!} }       | c =  }} {{eqn | r = \frac {n!} {\paren {\paren {n / 2}!}^2}       | c =  }} {{end-eqn}} So the middle term collapses to: {{begin-eqn}} {{eqn | l = R_n       | r = 2 \frac {n!} {\paren {\paren {n / 2}!}^2} \frac {e^{0 i \theta} } 2       | c =  }} {{eqn | r = \frac {n!} {\paren {\paren {n / 2}!}^2}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos^{2 n} \theta       | r = \paren {\frac {e^{i \theta} + e^{-i \theta} } 2}^{2 n}       | c = Cosine Exponential Formulation }} {{eqn | r = \frac 1 {2^{2 n} } \paren {e^{i \theta} + e^{-i \theta} }^{2 n}       | c = Exponent Combination Laws: Power of Product }} {{eqn | r = \frac 1 {2^{2 n} } \sum^{2 n}_{k \mathop = 0} \binom{2 n} k e^{k i \theta} e^{-\paren {2 n - k} i \theta}       | c = Binomial Theorem }} {{eqn | r = \frac 1 {2^{2 n} } \sum^{2 n}_{k \mathop = 0} \binom{2 n} k e^{-\paren {2 n - 2 k} i \theta}       | c = Exponential of Sum }} {{eqn | r = \frac 1 {2^{2 n} } \paren {\sum^{n \mathop - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \binom {2 n} n e^{\paren {2 n - 2 n} i \theta} + \sum^{2 n}_{k \mathop = n + 1} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} }       | c = partitioning the sum }} {{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n} } \paren {\sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \sum^{2 n}_{k \mathop = n + 1} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} }       | c = Exponential of Zero }} {{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n} } \paren {\sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \sum^{n - 1}_{k \mathop = 0} \binom {2 n} {2 n - k} e^{\paren {2 \paren {2 n - k} - 2 n} i \theta} }       | c = $k \mapsto 2 n - k$ }} {{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n} } \paren {\sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{-\paren {2 n - 2 k} i \theta} + \sum^{n - 1}_{k \mathop = 0} \binom {2 n} k e^{\paren {2 n - 2 k} i \theta} }       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \frac 1 {2^{2 n} } \binom {2 n} n + \frac 1 {2^{2 n - 1} } \sum^{n - 1}_{k \mathop = 0} \binom {2 n} k \map \cos {2 n - 2 k} \theta       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
Let $y \in \R$ such that: : $\exists x \in \R: x = \cot \left({y + \dfrac \pi 2}\right)$ Then: {{begin-eqn}} {{eqn | l = x       | r = \cot \left({y + \frac \pi 2}\right)       | c =  }} {{eqn | r = -\tan y       | c = Cotangent of Angle plus Right Angle }} {{eqn | r = \tan \left({-y}\right)       | c = Tangent Function is Odd }} {{end-eqn}} Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$. Then we can write $-y = \arctan x$. But then $\cot \left({y + \dfrac \pi 2}\right) = x$. Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$. Hence $y + \dfrac \pi 2 = \operatorname{arccot} x$. That is, $\dfrac \pi 2 = \operatorname{arccot} x + \arctan x$. {{qed}}
{{begin-eqn}} {{eqn | l = \map \arcsin {\frac 1 x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x       | r = \sin y       | c = {{Defof|Arcsine}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \csc y       | c = Cosecant is Reciprocal of Sine }} {{eqn | ll= \leadstoandfrom       | l = \arccsc x       | r = y       | c = {{Defof|Arccosecant}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arccos {\frac 1 x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x       | r = \cos y       | c = {{Defof|Arccosine}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \sec y       | c = Secant is Reciprocal of Cosine }} {{eqn | ll= \leadstoandfrom       | l = \arcsec x       | r = y       | c = {{Defof|Arcsecant}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arctan {\frac 1 x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 x       | r = \tan y       | c = {{Defof|Arctangent}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \cot y       | c = Cotangent is Reciprocal of Tangent }} {{eqn | ll= \leadstoandfrom       | l = \arccot x       | r = y       | c = {{Defof|Arccotangent}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arcsin {-x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = -x       | r = \sin y:       | rr = -\frac \pi 2 \le y \le \frac \pi 2       | c = {{Defof|Arcsine}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = -\sin y:       | rr = -\frac \pi 2 \le y \le \frac \pi 2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \map \sin {-y}:       | rr = -\frac \pi 2 \le y \le \frac \pi 2       | c = Sine Function is Odd }} {{eqn | ll= \leadstoandfrom       | l = \arcsin x       | r = -y       | c = {{Defof|Arcsine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arccos {-x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = -x       | r = \cos y:       | rr= 0 \le y \le \pi       | c = {{Defof|Arccosine}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = -\cos y:       | rr= -\pi \le y \le 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \map \cos {\pi - y}:       | rr= 0 \le y \le \pi       | c = Cosine of Supplementary Angle }} {{eqn | ll= \leadstoandfrom       | l = \arccos x       | r = \pi - y       | c = {{Defof|Arccosine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arctan {-x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = -x       | r = \tan y:       | rr= -\frac \pi 2 \le y \le \frac \pi 2       | c = {{Defof|Arctangent}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = -\tan y:       | rr= -\frac \pi 2 \le y \le \frac \pi 2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \map \tan {-y}:       | rr= -\frac \pi 2 \le y \le \frac \pi 2       | c = Tangent Function is Odd }} {{eqn | ll= \leadstoandfrom       | l = \arctan x       | r = -y       | c = {{Defof|Arctangent}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arccot {-x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = -x       | r = \cot y:       | rr= 0 \le y \le \pi       | c = {{Defof|Arccotangent}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = -\cot y:       | rr= -\pi \le y \le 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \map \cot {\pi - y}:       | rr= 0 \le y \le \pi       | c = Cotangent of Supplementary Angle }} {{eqn | ll= \leadstoandfrom       | l = \arccot x       | r = \pi - y       | c = {{Defof|Arccotangent}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arcsec {-x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = -x       | r = \sec y:       | rr= 0 \le y \le \pi       | c = {{Defof|Arcsecant}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = -\sec y:       | rr= -\pi \le y \le 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \map \sec {\pi - y}:       | rr= 0 \le y \le \pi       | c = Cosecant of Supplementary Angle }} {{eqn | ll= \leadstoandfrom       | l = \arcsec x       | r = \pi - y       | c = {{Defof|Arcsecant}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \arccsc {-x}       | r = y       | c =  }} {{eqn | ll= \leadstoandfrom       | l = -x       | r = \csc y:       | rr= -\frac \pi 2 \le y \le \frac \pi 2       | c = {{Defof|Arccosecant}} }} {{eqn | ll= \leadstoandfrom       | l = x       | r = -\csc y:       | rr= -\frac \pi 2 \le y \le \frac \pi 2       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | r = \map \csc {-y}:       | rr= -\frac \pi 2 \le y \le \frac \pi 2       | c = Cosecant Function is Odd }} {{eqn | ll= \leadstoandfrom       | l = \arccsc x       | r = -y       | c = {{Defof|Arccosecant}} }} {{end-eqn}} {{qed}}
Let $d = \dfrac a {\sin A}$. From the Law of Sines, let: :$d = \dfrac a {\sin A} = \dfrac b {\sin B}$ so that: {{begin-eqn}} {{eqn | l = a       | r = d \sin A       | c =  }} {{eqn | l = b       | r = d \sin B       | c =  }} {{eqn | ll= \leadsto       | l = \frac {a + b} {a - b}       | r = \frac {d \sin A + d \sin B} {d \sin A - d \sin B}       | c =  }} {{eqn | r = \frac {\sin A + \sin B} {\sin A - \sin B}       | c =  }} {{eqn-intertext|Let $C {{=}} \frac 1 2 \paren {A + B}$ and $D {{=}} \frac 1 2 \paren {A - B}$, and proceed as follows:}} {{eqn | ll= \leadsto       | l = \frac {a + b} {a - b}       | r = \frac {2 \sin C \cos D} {\sin A - \sin B}       | c = Prosthaphaeresis Formula for Sine plus Sine }} {{eqn | r = \frac {2 \sin C \cos D} {2 \sin D \cos C}       | c = Prosthaphaeresis Formula for Sine minus Sine }} {{eqn | r = \frac {\frac {\sin C} {\cos C} } {\frac {\sin D} {\cos D} }       | c = dividing top and bottom by $\cos C \cos D$ }} {{eqn | r = \frac {\tan C} {\tan D}       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }       | c = substituting back for $C$ and $D$ }} {{end-eqn}} {{qed}}
Let $Q$ be the area of $\triangle ABC$. From Area of Triangle in Terms of Two Sides and Angle: :$Q = \dfrac {b c \sin A} 2$ From Heron's Formula: :$Q = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ Equating the two: :$\dfrac {b c \sin A} 2 = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$ from which follows the result. {{qed}}
From Functional Completeness over Finite Number of Arguments, it suffices to consider binary truth functions. From Count of Truth Functions, there are $16$ of these. These are enumerated in Binary Truth Functions, and are analysed in turn as follows. === Constant Functions === There are two constant functions: {{begin-eqn}} {{eqn | l = f_F \left({p, q}\right)       | r = F }} {{eqn | l = f_T \left({p, q}\right)       | r = T }} {{end-eqn}} From Biconditional Properties and Exclusive Or Properties: {{begin-eqn}} {{eqn | l = p \iff p       | o = \dashv \vdash       | r = \top \dashv \vdash f_T \left({p, q}\right) }} {{eqn | l = p \oplus p       | o = \dashv \vdash       | r = \bot \dashv \vdash f_F \left({p, q}\right) }} {{end-eqn}} where $\oplus$ and $\iff$ denote exclusive or and biconditional respectively. So both of the constant functions can be expressed in terms of $\oplus$ and $\iff$. {{qed|lemma}} === Equivalence and Non-Equivalence === By definition of exclusive or: : $p \oplus q \dashv \vdash \neg \left({p \iff q}\right)$. Thus $\oplus$ can be expressed in terms of $\neg$ and $\iff$. By definition of biconditional: : $p \iff q \dashv \vdash \left({p \implies q}\right) \land \left({q \implies p}\right)$ Thus $\iff$ can be expressed in terms of $\implies$ and $\land$. {{qed|lemma}} === Projections and Negated Projections === There are two projections: : $\operatorname{pr}_1 \left({p, q}\right) = p$ : $\operatorname{pr}_2 \left({p, q}\right) = q$ We note that: : $p \land p \dashv \vdash p \dashv \vdash \operatorname{pr}_1 \left({p, q}\right)$ : $p \lor p \dashv \vdash p \dashv \vdash \operatorname{pr}_1 \left({p, q}\right)$ and similarly for $\operatorname{pr}_2 \left({p, q}\right)$. So the projections can be expressed in terms of either $\land$ or $\lor$. There are two negated projections: : $\overline {\operatorname{pr}_1} \left({p, q}\right) = \neg p$ : $\overline {\operatorname{pr}_2} \left({p, q}\right) = \neg q$ It immediately follows from the above that these can be expressed in terms of either: : $\neg$ and $\land$ or: : $\neg$ and $\lor$ {{qed|lemma}} === NAND and NOR === There are the NAND and NOR operators: : $p \uparrow q$ : $p \downarrow q$ By definition of NAND: : $p \uparrow q \dashv \vdash \neg \left({p \land q}\right)$ By definition of NOR: : $p \downarrow q \dashv \vdash \neg \left({p \lor q}\right)$ So: : $\uparrow$ can be expressed in terms of $\neg$ and $\land$ : $\downarrow$ can be expressed in terms of $\neg$ and $\lor$ {{qed|lemma}} === Conjunction, Disjunction, Conditional === There are the conjunction and disjunction operators: : $p \land q$ : $p \lor q$ There are two conditionals: : $p \implies q$ : $q \implies p$ There are two  negated conditionals: : $\neg \left({p \implies q}\right)$ : $\neg \left({q \implies p}\right)$ All of the above are already expressed in terms of $\neg, \land, \lor, \implies$. {{qed|lemma}} Thus all sixteen of the Binary Truth Functions can be expressed in terms of: : $\neg, \land, \lor, \implies$ That is: : $\left\{{\neg, \land, \lor, \implies}\right\}$ is functionally complete. {{qed}} Category:Functional Completeness 0qv4fsb1giyp2qwmxxefz3707f3rl8b
From Functionally Complete Logical Connectives: Negation and Conjunction, $\set {\neg, \land}$ is functionally complete. That is: any expression can be expressed in terms of $\neg$ and $\land$. From De Morgan's laws: Conjunction, we have that: :$p \land q \dashv \vdash \neg \paren {\neg p \lor \neg q}$ Thus all occurrences of $\land$ can be replaced by $\lor$ and $\neg$. Thus any expression can be expressed in terms of $\neg$ and $\lor$. That is: $\set {\neg, \lor}$ is functionally complete. {{qed}}
From Functionally Complete Logical Connectives: Negation, Conjunction, Disjunction and Implication, all sixteen of the binary truth functions can be expressed in terms of $\neg, \land, \lor, \implies$. From Conjunction and Implication, we have that: :$p \implies q \dashv \vdash \neg \paren {p \land \neg q}$ From De Morgan's laws: Disjunction, we have that: :$p \lor q \dashv \vdash \neg \paren {\neg p \land \neg q}$ So any instance of either $\implies$ or $\lor$ can be replaced identically with one using just $\neg$ and $\land$. It follows that $\set {\neg, \land}$ is functionally complete. {{qed}}
From Functionally Complete Logical Connectives: Negation and Conjunction, we can represent any boolean expression in terms of $\land$ and $\neg$. From Conjunction and Implication, we have that: :$p \land q \dashv \vdash \neg \paren {p \implies \neg q}$ So it follows that we can replace all occurrences of $\land$ by $\implies$ and $\neg$. Thus $\set {\neg, \implies}$ is functionally complete. {{qed}}
From the stronger results: :Functionally Complete Logical Connectives: Negation and Disjunction: ::the set of logical connectives: $\set {\neg, \lor}$ is functionally complete :Functionally Complete Logical Connectives: Negation and Conjunction: ::the set of logical connectives: $\set {\neg, \land}$ is functionally complete it follows directly that $\set {\neg, \land, \lor}$ is likewise functionally complete. {{qed}} Category:Functional Completeness eh8et9jrx7lv8b1gvg1ipjzycdrvr03
From Functionally Complete Logical Connectives: Negation and Conjunction, any boolean expression can be expressed in terms of $\land$ and $\neg$. From NAND with Equal Arguments: :$\neg p \dashv \vdash p \uparrow p$ From Conjunction in terms of NAND: :$p \land q \dashv \vdash \paren {p \uparrow q} \uparrow \paren {p \uparrow q}$ demonstrating that $p \land q$ is expressed solely in terms of $\uparrow$. Thus any boolean expression can be represented solely in terms of $\uparrow$. That is, $\set {\uparrow}$ is functionally complete. {{qed}}
From Functionally Complete Logical Connectives: Negation and Disjunction, any boolean expression can be expressed in terms of $\lor$ and $\neg$. From NOR with Equal Arguments: :$\neg p \dashv \vdash p \downarrow p$ From Disjunction in terms of NOR: :$p \lor q \dashv \vdash \paren {p \downarrow q} \downarrow \paren {p \downarrow q}$ demonstrating that $p \lor q$ can be represented solely in terms of $\downarrow$. That is, $\set {\downarrow}$ is functionally complete. {{qed}}
Let $\uparrow$ and $\downarrow$ denote NAND and NOR respectively. From: : NAND is Functionally Complete and : NOR is Functionally Complete the singleton sets $\left\{{\uparrow}\right\}$ and $\left\{{\downarrow}\right\}$ are functionally complete. Suppose $\circ$ is a binary logical connective such that $\left\{{\circ}\right\}$ is a functionally complete set. The unary logical connective $\neg$ has to be equivalent to some formula: : $\cdots \circ \left({p \circ p}\right) \circ \cdots$ Suppose some interpretation $v$ assigns $T$ to $p$. Then $v \left({\neg p}\right) = F$ So: : $v \left({\cdots \circ \left({p \circ p}\right) \circ \cdots}\right) = F$ So it has to be true that $T \circ T = F$, otherwise: : $v \left({\cdots \circ \left({T \circ T}\right) \circ \cdots}\right) = v \left({\cdots \circ \left({T}\right) \circ \cdots}\right) = T$ Similarly: :$F \circ F = T$ Now consider $T \circ F$ and $F \circ T$. Suppose $T \circ F = F$. If $F \circ T = T$, then we have the function defined by this truth table: {| border="1" cellpadding="4" cellspacing="1" style="text-align:center;" |-  ! $v \left({p}\right)$ ! $v \left({q}\right)$ ! $v \left({p \circ q}\right)$ |- | $F$ || $F$ || $T$ |- | $F$ || $T$ || $T$ |- | $T$ || $F$ || $F$ |- | $T$ || $T$ || $F$ |} which is $\neg p$, and hence only negation would be definable. So if $T \circ F = F$ we need $F \circ T = F$. This gives the truth table for the NOR function: {| border="1" cellpadding="4" cellspacing="1" style="text-align:center;" |-  ! $v \left({p}\right)$ ! $v \left({q}\right)$ ! $v \left({p \circ q}\right)$ |- | $F$ || $F$ || $T$ |- | $F$ || $T$ || $F$ |- | $T$ || $F$ || $F$ |- | $T$ || $T$ || $F$ |} ... which we have seen is functionally complete. Similarly, Suppose $T \circ F = T$. If $F \circ T = F$, then we have the function defined by this truth table: {| border="1" cellpadding="4" cellspacing="1" style="text-align:center;" |-  ! $v \left({p}\right)$ ! $v \left({q}\right)$ ! $v \left({p \circ q}\right)$ |- | $F$ || $F$ || $T$ |- | $F$ || $T$ || $F$ |- | $T$ || $F$ || $T$ |- | $T$ || $T$ || $F$ |} which is $\neg q$, and hence only negation would be definable. So if $T \circ F = T$ we need $F \circ T = T$. This gives the truth table for the NAND function: {| border="1" cellpadding="4" cellspacing="1" style="text-align:center;" |-  ! $v \left({p}\right)$ ! $v \left({q}\right)$ ! $v \left({p \circ q}\right)$ |- | $F$ || $F$ || $T$ |- | $F$ || $T$ || $T$ |- | $T$ || $F$ || $T$ |- | $T$ || $T$ || $F$ |} ... which we have seen is functionally complete. Thus it follows that there can be no functionally complete binary logical connectives apart from NAND and NOR. {{qed}}
We apply the Method of Truth Tables to the proposition: :$\paren {p \implies q} \iff \paren {q \implies p}$ $\begin{array}{|ccc|c|ccc|} \hline p & \implies & q) & \iff & (q & \implies & p) \\ \hline F & T & F & T & F & T & F \\ F & T & T & F & T & F & F \\ T & F & F & F & F & T & T \\ T & T & T & T & T & T & T \\ \hline \end{array}$ As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations. {{qed}}
{{BeginTableau|\forall x: \map P x \vdash \neg \exists x: \neg \map P x}} {{Premise|1|\forall x: \map P x}} {{Assumption|2|\exists x: \neg \map P x}} {{TableauLine|n = 3|pool = 2|f = \neg \map P {\mathbf a}|rlnk = Existential Instantiation|rtxt = Existential Instantiation|dep = 2|c = for some arbitrary $\mathbf a$}} {{TableauLine|n = 4|pool = 1|f = \map P {\mathbf a}|rlnk = Universal Instantiation|rtxt = Universal Instantiation|dep = 1}} {{NonContradiction|5|1, 2|3|4}} {{Contradiction|6|1|\neg \exists x: \neg \map P x|2|5}} {{EndTableau|lemma}} {{BeginTableau|\neg \exists x: \neg \map P x \vdash \forall x: \map P x}} {{Premise|1|\neg \exists x: \neg \map P x}} {{Assumption|2|\neg \forall x: \map P x}} {{SequentIntro|3|2|\exists x: \neg \map P x|2|Denial of Universality}} {{NonContradiction|4|1, 2|1|3}} {{Reductio|5|1|\forall x: \map P x|2|4}} {{EndTableau|qed}} {{LEM|Reductio ad Absurdum}} {{Namedfor|Augustus De Morgan|cat = De Morgan}}
{{BeginTableau|\neg \forall x: \neg \map P x \vdash \exists x: \map P x}} {{Premise|1|\neg \forall x: \neg \map P x}} {{Assumption|2|\neg \exists x: \map P x}} {{SequentIntro|3|2|\forall x: \neg \map P x|2|Denial of Existence}} {{NonContradiction|4|1, 2|1|3}} {{Reductio|5|1|\exists x: \map P x|2|4}} {{EndTableau|lemma}} {{BeginTableau|\exists x: \map P x \vdash \neg \forall x: \neg \map P x}} {{Premise|1|\exists x: \map P x}} {{Assumption|2|\forall x: \neg \map P x}} {{TableauLine|n = 3|pool = 1|f = \map P {\mathbf a}|rlnk = Existential Instantiation|rtxt = Existential Instantiation|dep = 1|for some arbitrary $\mathbf a$}} {{TableauLine|n = 4|pool = 2|f = \neg \map P {\mathbf a}|rlnk = Universal Instantiation|rtxt = Universal Instantiation|dep = 2}} {{NonContradiction|5|1, 2|3|4}} {{Contradiction|6|1|\neg \forall x: \neg \map P x|2|5}} {{EndTableau|qed}} {{LEM|Reductio ad Absurdum}} {{Namedfor|Augustus De Morgan|cat = De Morgan}}
{{begin-eqn}} {{eqn | l = p \land q       | o = \dashv \vdash       | r = \neg \neg \paren {p \land q}       | c = Double Negation }} {{eqn | o = \dashv \vdash       | r = \neg \paren {p \uparrow q}       | c = {{Defof|Logical NAND}} }} {{eqn | o = \dashv \vdash       | r = \paren {p \uparrow q} \uparrow \paren {p \uparrow q}       | c = NAND with Equal Arguments }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = p \lor q       | o = \dashv \vdash       | r = \neg \neg \paren {p \lor q}       | c = Double Negation }} {{eqn | o = \dashv \vdash       | r = \neg \paren {p \downarrow q}       | c = {{Defof|Logical NOR}} }} {{eqn | o = \dashv \vdash       | r = \paren {p \downarrow q} \downarrow \paren {p \downarrow q}       | c = NOR with Equal Arguments }} {{end-eqn}} {{qed}} Category:Logical NOR Category:Disjunction 4dw5e72qt5juo7ko2pxvsrci6p0nyyc
{{begin-eqn}} {{eqn | l = p \lor q       | o = \dashv \vdash       | r = \neg \paren {\neg p \land \neg q}       | c = De Morgan's Laws (Logic): Disjunction }} {{eqn | o = \dashv \vdash       | r = \neg p \uparrow \neg q       | c = {{Defof|Logical NAND}} }} {{eqn | o = \dashv \vdash       | r = \paren {p \uparrow p} \uparrow \paren {q \uparrow q}       | c = NAND with Equal Arguments }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = p \implies q       | o = \dashv \vdash       | r = \neg \paren {p \land \neg q}       | c = Implication Equivalent to Negation of Conjunction with Negative }} {{eqn | o = \dashv \vdash       | r = p \uparrow \neg q       | c = Definition of NAND }} {{eqn | o = \dashv \vdash       | r = p \uparrow \paren {q \uparrow q}       | c = NAND with Equal Arguments }} {{end-eqn}} {{qed}}