Principle of Commutation/Reverse Implication/Formulation 2

Theorem

$\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$


Proof

By the tableau method of natural deduction:

$\vdash \paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} } $
Line Pool Formula Rule Depends upon Notes
1 1 $q \implies \paren {p \implies r}$ Assumption (None)
2 1 $p \implies \paren {q \implies r}$ Sequent Introduction 1 Principle of Commutation: Reverse Implication: Formulation 1
3 $\paren {q \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ Rule of Implication: $\implies \II$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$

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