Biconditional as Disjunction of Conjunctions/Formulation 1/Forward Implication

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Theorem

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


Proof

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: $\iff \mathcal E_1$ 1
3 1 $q \implies p$ Biconditional Elimination: $\iff \mathcal E_2$ 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: $\land \mathcal I$ 5, 6
8 1, 5 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Rule of Addition: $\lor \mathcal I_1$ 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: $\land \mathcal I$ 9, 10
12 1, 9 $\left({p \land q}\right) \lor \left({\neg p \land \neg q}\right)$ Rule of Addition: $\lor \mathcal I_2$ 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

$\blacksquare$

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.