Biconditional is Commutative/Formulation 2
Theorems
The biconditional operator is commutative:
- $\vdash \paren {p \iff q} \iff \paren {q \iff p}$
Proof
By the tableau method of natural deduction:
| Line | Pool | Formula | Rule | Depends upon | Notes | |
|---|---|---|---|---|---|---|
| 1 | 1 | $\paren {p \iff q}$ | Assumption | (None) | ||
| 2 | 1 | $\paren {q \iff p}$ | Sequent Introduction | 1 | Biconditional is Commutative: Formulation 1 | |
| 3 | 1 | $\paren {p \iff q} \implies \paren {q \iff p}$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged | |
| 4 | 4 | $\paren {p \iff q}$ | Assumption | (None) | ||
| 5 | 4 | $\paren {q \iff p}$ | Sequent Introduction | 4 | Biconditional is Commutative: Formulation 1 | |
| 6 | 4 | $\paren {p \iff q} \implies \paren {q \iff p}$ | Rule of Implication: $\implies \II$ | 4 – 5 | Assumption 4 has been discharged | |
| 7 | $\paren {p \iff q} \iff \paren {q \iff p}$ | Biconditional Introduction: $\iff \II$ | 3, 6 |
$\blacksquare$
Sources
- 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T92}$