Principle of Dilemma/Formulation 2/Forward Implication/Proof 1

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Theorem

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


Proof

By the tableau method of natural deduction:

$\vdash \paren {p \implies q} \land \paren {\neg p \implies q} \implies q$
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \land \paren {\neg p \implies q}$ Assumption (None)
2 1 $q$ Sequent Introduction 1 Principle of Dilemma: Formulation 1
3 $\paren {p \implies q} \land \paren {\neg p \implies q} \implies q$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$