Negation implies Negation of Conjunction/Case 2

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Theorem

$\neg q \implies \neg \left({p \land q}\right)$


Proof

By the tableau method of natural deduction:

$\neg q \implies \neg \left({p \land q}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\neg q$ Assumption (None)
2 2 $p \land q$ Assumption (None)
3 2 $q$ Rule of Simplification: $\land \mathcal E_2$ 2
4 1, 2 $\bot$ Principle of Non-Contradiction: $\neg \mathcal E$ 3, 1
5 1 $\neg \left({p \land q}\right)$ Proof by Contradiction: $\neg \mathcal I$ 2 – 4 Assumption 2 has been discharged
6 $\neg q \implies \neg \left({p \land q}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 5 Assumption 1 has been discharged

$\blacksquare$


Sources