Biconditional Elimination/Sequent Form/Proof 1

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:

$p \iff q \vdash p \implies q$
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:

$p \iff q \vdash q \implies p$
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

$\blacksquare$

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