Rule of Material Equivalence/Formulation 2/Proof 1

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Theorem

$\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} }$


Proof

By the tableau method of natural deduction:

$\vdash \paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} } $
Line Pool Formula Rule Depends upon Notes
1 1 $p \iff q$ Assumption (None)
2 1 $\paren {\paren {p \implies q} \land \paren {q \implies p} }$ Sequent Introduction 1 Rule of Material Equivalence: Formulation 1
3 $\paren {p \iff q} \implies \paren {\paren {p \implies q} \land \paren {q \implies p} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged
4 4 $\paren {p \implies q} \land \paren {q \implies p}$ Assumption (None)
5 4 $p \iff q$ Sequent Introduction 4 Rule of Material Equivalence: Formulation 1
6 $\paren {\paren {p \implies q} \land \paren {q \implies p} } \implies \paren {p \iff q}$ Rule of Implication: $\implies \II$ 4 – 5 Assumption 4 has been discharged
7 $\paren {p \iff q} \iff \paren {\paren {p \implies q} \land \paren {q \implies p} }$ Biconditional Introduction: $\iff \II$ 3, 6

$\blacksquare$