Biconditional as Disjunction of Conjunctions/Formulation 1/Forward Implication

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$p \iff q \vdash \left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$


By the tableau method of natural deduction:

$p \iff q \vdash \left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 1 $p \implies q$ Biconditional Elimination: $\iff \EE_1$ 1
3 1 $q \implies p$ Biconditional Elimination: $\iff \EE_2$ 1
4 $p \lor \neg p$ Law of Excluded Middle (None)
5 5 $p$ Assumption (None)
6 1, 5 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 5
7 1, 5 $p \land q$ Rule of Conjunction: $\land \II$ 5, 6
8 1, 5 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Rule of Addition: $\lor \II_1$ 7
9 9 $\neg p$ Assumption (None)
10 1, 9 $\neg q$ Modus Tollendo Tollens (MTT) 3, 9
11 1, 9 $\neg p \land \neg q$ Rule of Conjunction: $\land \II$ 9, 10
12 1, 9 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Rule of Addition: $\lor \II_2$ 11
13 1 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Proof by Cases: $\text{PBC}$ 1, 5 – 8, 9 – 12 Assumptions 5 and 9 have been discharged


Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom.

This in turn invalidates this theorem from an intuitionistic perspective.