Biconditional Elimination/Sequent Form/Proof 1/Form 1
Jump to navigation
Jump to search
Theorem
- $p \iff q \vdash p \implies q$
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 |
$\blacksquare$