Non-Equivalence as Conjunction of Disjunction with Negation of Conjunction

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Theorem

$\neg \paren {p \iff q} \dashv \vdash \paren {p \lor q} \land \neg \paren {p \land q}$

That is, negation of the biconditional means the same thing as either-or but not both, that is, exclusive or.


Proof 1

By the tableau method of natural deduction:

$\neg \paren {p \iff q} \vdash \paren {p \lor q} \land \neg \paren {p \land q} $
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \iff q}$ Premise (None)
2 1 $\paren {\neg p \land q} \lor \paren {p \land \neg q}$ Sequent Introduction 1 Non-Equivalence as Disjunction of Conjunctions: Formulation 1
3 1 $\paren {p \land \neg q} \lor \paren {\neg p \land q}$ Sequent Introduction 2 Disjunction is Commutative
4 1 $\paren {p \land \neg q} \lor \paren {q \land \neg p}$ Sequent Introduction 3 Conjunction is Commutative
5 1 $\paren {\paren {p \lor q} \land \neg q} \lor \paren {\paren {q \lor p} \land \neg p}$ Sequent Introduction 4 Conjunction of Disjunction with Negation is Conjunction with Negation
6 1 $\paren {\paren {p \lor q} \land \neg q} \lor \paren {\paren {p \lor q} \land \neg p}$ Sequent Introduction 5 Disjunction is Commutative
7 1 $\paren {p \lor q} \land \paren {\neg q \lor \neg p}$ Sequent Introduction 6 Conjunction Distributes over Disjunction
8 1 $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ Sequent Introduction 7 Disjunction is Commutative
9 1 $\paren {p \lor q} \land \neg \paren {\neg \neg p \land \neg \neg q}$ Sequent Introduction 8 De Morgan's Laws: Disjunction
10 1 $\paren {p \lor q} \land \neg \paren {p \land q}$ Double Negation Elimination: $\neg \neg \EE$ 9


By the tableau method of natural deduction:

$\paren {p \lor q} \land \neg \paren {p \land q} \vdash \neg \paren {p \iff q} $
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \lor q} \land \neg \paren {p \land q}$ Premise (None)
2 1 $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ Sequent Introduction 1 De Morgan's Laws: Disjunction of Negations
3 1 $\paren {p \lor q} \land \paren {\neg p \lor \neg q}$ Sequent Introduction 2 Disjunction is Commutative
4 1 $\paren {\paren {p \lor q} \land \neg q} \lor \paren {\paren {p \lor q} \land \neg p}$ Sequent Introduction 3 Conjunction Distributes over Disjunction
5 1 $\paren {p \land \neg q} \lor \paren {q \land \neg p}$ Sequent Introduction 4 Conjunction of Disjunction with Negation is Conjunction with Negation
6 1 $\paren {q \land \neg p} \lor \paren {p \land \neg q}$ Sequent Introduction 5 Disjunction is Commutative
7 1 $\paren {\neg p \land q} \lor \paren {p \land \neg q}$ Sequent Introduction 6 Conjunction is Commutative
8 1 $\neg \paren {p \iff q}$ Sequent Introduction 6 Non-Equivalence as Disjunction of Conjunctions: Formulation 1

$\blacksquare$


Proof by Truth Table

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.


$\begin{array}{|cccc||cccccccc|} \hline \neg & (p & \iff & q) & (p & \lor & q) & \land & \neg & (p & \land & q) \\ \hline \F & \F & \T & \F & \F & \F & \F & \F & \T & \F & \F & \F \\ \T & \F & \F & \T & \F & \T & \T & \T & \T & \F & \F & \T \\ \T & \T & \F & \F & \T & \T & \F & \T & \T & \T & \F & \F \\ \F & \T & \T & \T & \T & \T & \T & \F & \F & \T & \T & \T \\ \hline \end{array}$

$\blacksquare$