Biconditional is Commutative/Formulation 1/Proof 2
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Theorem
- $p \iff q \dashv \vdash q \iff p$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \iff q$ | Premise | (None) | ||
2 | 1 | $\paren {p \implies q} \land \paren {q \implies p}$ | Sequent Introduction | 1 | Rule of Material Equivalence | |
3 | 1 | $\paren {q \implies p} \land \paren {p \implies q}$ | Sequent Introduction | 2 | Conjunction is Commutative | |
4 | 1 | $q \iff p$ | Sequent Introduction | 3 | Rule of Material Equivalence |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $q \iff p$ | Premise | (None) | ||
2 | 1 | $\paren {q \implies p} \land \paren {p \implies q}$ | Sequent Introduction | 1 | Rule of Material Equivalence | |
3 | 1 | $\paren {p \implies q} \land \paren {q \implies p}$ | Sequent Introduction | 2 | Conjunction is Commutative | |
4 | 1 | $p \iff q$ | Sequent Introduction | 3 | Rule of Material Equivalence |
$\blacksquare$
Sources
- 1965: E.J. Lemmon: Beginning Logic ... (previous) ... (next): Chapter $1$: The Propositional Calculus $1$: $4$ The Biconditional: Theorem $24$