Modus Ponendo Tollens/Sequent Form/Case 1

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Theorem

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


Proof

By the tableau method of natural deduction:

$\neg \paren {p \land q}, p \vdash \neg q$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg \paren {p \land q}$ Premise (None)
2 2 $p$ Premise (None)
3 3 $q$ Assumption (None)
4 2, 3 $p \land q$ Rule of Conjunction: $\land \II$ 2, 3
5 1, 2, 3 $\bot$ Principle of Non-Contradiction: $\neg \EE$ 4, 1
6 1, 2 $\neg q$ Proof by Contradiction: $\neg \II$ 3 – 5 Assumption 3 has been discharged

$\blacksquare$


Sources