Modus Ponendo Ponens/Variant 3/Proof 1
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Theorem
- $\vdash \paren {\paren {p \implies q} \land p} \implies q$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \implies q} \land p$ | Assumption | (None) | ||
2 | 1 | $p \implies q$ | Rule of Simplification: $\land \EE_1$ | 1 | ||
3 | 1 | $p$ | Rule of Simplification: $\land \EE_2$ | 1 | ||
4 | 1 | $q$ | Modus Ponendo Ponens: $\implies \mathcal E$ | 2, 3 | ||
5 | $\paren {\paren {p \implies q} \land p} \implies q$ | Rule of Implication: $\implies \II$ | 1 – 4 | Assumption 1 has been discharged |
$\blacksquare$