Biconditional is Commutative/Formulation 1/Proof 1
Jump to navigation
Jump to search
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 | $p \implies q$ | Biconditional Elimination: $\iff \EE_1$ | 1 | ||
3 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \EE_2$ | 1 | ||
4 | 1 | $q \iff p$ | Biconditional Introduction: $\iff \II$ | 3, 2 |
$\Box$
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $q \iff p$ | Premise | (None) | ||
2 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \EE_1$ | 1 | ||
3 | 1 | $p \implies q$ | Biconditional Elimination: $\iff \EE_2$ | 1 | ||
4 | 1 | $p \iff q$ | Biconditional Introduction: $\iff \II$ | 3, 2 |
$\blacksquare$