Changes

==Exclusive disjunction==

==Exclusive disjunction==

''[[Exclusive disjunction]]'', also known as ''logical inequality'' or ''symmetric difference'', is an [[logical operation|operation]] on two [[logical value]]s, typically the values of two [[proposition]]s, that produces a value of ''true'' just in case exactly one of its operands is true.

'''[[Exclusive disjunction]]''', also known as '''logical inequality''' or '''symmetric difference''', is an operation on two logical values, typically the values of two propositions, that produces a value of ''true'' just in case exactly one of its operands is true.

The truth table of '''p XOR q''' (also written as '''p + q''', '''p ⊕ q''', or '''p ≠ q''') is as follows:

The truth table of <math>p ~\operatorname{XOR}~ q,</math> also written <math>p + q\!</math> or <math>p \ne q,\!</math> appears below:

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<br>

{| align="center" border="1" cellpadding="8" cellspacing="0" style="background:#f8f8ff; font-weight:bold; text-align:center; width:45%"

{| align="center" border="1" cellpadding="8" cellspacing="0" style="text-align:center; width:45%"

|+ '''Exclusive Disjunction'''

|+ style="height:30px" | <math>\text{Exclusive Disjunction}\!</math>

|- style="background:#e6e6ff"

|- style="height:40px; background:#f0f0ff"

! style="width:15%" | p

| style="width:33%" | <math>p\!</math>

! style="width:15%" | q

| style="width:33%" | <math>q\!</math>

! style="width:15%" | p XOR q

| style="width:33%" | <math>p ~\operatorname{XOR}~ q</math>

|-

|-

| F || F || F

| <math>\operatorname{F}</math> || <math>\operatorname{F}</math> || <math>\operatorname{F}</math>

|-

|-

| F || T || T

| <math>\operatorname{F}</math> || <math>\operatorname{T}</math> || <math>\operatorname{T}</math>

|-

|-

| T || F || T

| <math>\operatorname{T}</math> || <math>\operatorname{F}</math> || <math>\operatorname{T}</math>

|-

|-

| T || T || F

| <math>\operatorname{T}</math> || <math>\operatorname{T}</math> || <math>\operatorname{F}</math>

|}

|}

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<br>

The following equivalents can then be deduced:

The following equivalents may then be deduced:

: <math>\begin{matrix}

p + q & = & (p \land \lnot q) & \lor & (\lnot p \land q) \\

\\

<math>\begin{matrix}

       & = & (p \lor q) & \land & (\lnot p \lor \lnot q) \\

p + q & = & (p \land \lnot q) & \lor & (\lnot p \land q)

\\

\\[6pt]

       & = & (p \lor q) & \land & (\lnot p \lor \lnot q)

\\[6pt]

       & = & (p \lor q) & \land & \lnot (p \land q)

       & = & (p \lor q) & \land & \lnot (p \land q)

\end{matrix}</math>

\end{matrix}</math>

==Logical implication==

==Logical implication==