Principle of Dilemma/Formulation 1/Reverse Implication

Theorem

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


Proof

By the tableau method of natural deduction:

$q \vdash \paren {p \implies q} \land \paren {\neg p \implies q} $
Line Pool Formula Rule Depends upon Notes
1 1 $q$ Premise (None)
2 1 $p \implies q$ Sequent Introduction 1 True Statement is implied by Every Statement
3 1 $\neg p \implies q$ Sequent Introduction 1 True Statement is implied by Every Statement
4 1 $\paren {p \implies q} \land \paren {\neg p \implies q}$ Rule of Conjunction: $\land \II$ 2, 3

$\blacksquare$

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