Biconditional is Commutative/Formulation 1/Proof 1

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Theorem

$p \iff q \dashv \vdash q \iff p$


Proof

By the tableau method of natural deduction:

$p \iff q \vdash q \iff p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Premise (None)
2 1 $p \implies q$ Biconditional Elimination: $\iff \mathcal E_1$ 1
3 1 $q \implies p$ Biconditional Elimination: $\iff \mathcal E_2$ 1
4 1 $q \iff p$ Biconditional Introduction: $\iff \mathcal I$ 3, 2

$\Box$


By the tableau method of natural deduction:

$q \iff p \vdash p \iff q$
Line Pool Formula Rule Depends upon Notes
1 1 $q \iff p$ Premise (None)
2 1 $q \implies p$ Biconditional Elimination: $\iff \mathcal E_1$ 1
3 1 $p \implies q$ Biconditional Elimination: $\iff \mathcal E_2$ 1
4 1 $p \iff q$ Biconditional Introduction: $\iff \mathcal I$ 3, 2

$\blacksquare$