Biconditional Elimination/Sequent Form/Proof 1
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Theorem
\(\text {(1)}: \quad\) | \(\ds p \iff q\) | \(\vdash\) | \(\ds p \implies q\) | |||||||||||
\(\text {(2)}: \quad\) | \(\ds p \iff q\) | \(\vdash\) | \(\ds q \implies p\) |
Proof
Form 1
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 |
$\blacksquare$
Form 2
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $p \iff q$ | Premise | (None) | ||
2 | 1 | $q \implies p$ | Biconditional Elimination: $\iff \EE_2$ | 1 |
$\blacksquare$