Modus Ponendo Ponens/Variant 2/Proof 1

Theorem

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


Proof

By the tableau method of natural deduction:

$\vdash p \implies \paren {\paren {p \implies q} \implies q} $
Line Pool Formula Rule Depends upon Notes
1 1 $p$ Assumption (None)
2 2 $p \implies q$ Assumption (None)
3 1, 2 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 1
4 1 $\paren {p \implies q} \implies q$ Rule of Implication: $\implies \II$ 2 – 3 Assumption 2 has been discharged
5 $p \implies \paren {\paren {p \implies q} \implies q}$ Rule of Implication: $\implies \II$ 1 – 4 Assumption 1 has been discharged

$\blacksquare$


Sources

  • 1964: Donald Kalish and Richard Montague: Logic: Techniques of Formal Reasoning ... (previous) ... (next): $\text{I}$: 'NOT' and 'IF': $\S 5$: Theorem $\text{T3}$
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