Question

Justify the rule of universal modus tollens by showing that the premises\(\forall x(P(x) \to Q(x))\) and\(\neg Q(a)\) for a particular element a in the domain, imply\(\neg P(a)\).

Short answer

The premises \(\forall x(P(x) \to Q(x))\) and\(\neg Q(a)\) for a particular element a in the domain implies\(\neg P(a)\).

Step-by-step solution

Rules of Inference

• Modus Ponens: \(\begin{array}{l}(p \wedge (p \to q)) \to q\\(P(a) \wedge (P(a) \to Q(a)) \to Q(a)\end{array}\)

• Modus Tollens: \((\neg q \wedge (p \to q)) \to \neg p\)

• Universal instantiation:It is a rule of inference which used to conclude that P(a) is true, where a is a particular member of the domain, with the given premise\(\forall xP(x)\).

Proof

Given: Premises\(\forall x(P(x) \to Q(x))\)and\(\neg Q(a)\)for a particular element a in the domain.

To show:\(\neg P(a)\)is true.

Proof: UsingUniversal Instantiationrule, \(P(a) \to Q(a)\) is true, where a is a particular member of the domain since it is given that \(\forall x(P(x) \to Q(x))\).

Now, the premises are\(\neg P(a)\)and\(\neg Q(a)\)and the conclusion should be\(\neg P(a)\).

Let\(\neg P(a)\)is not true.

\(\neg P(a)\)is true.

؞By modus ponens, \((P(a) \wedge (P(a) \to Q(a)) \to Q(a)\)

But this contradicts the premise\(\neg Q(a)\)

So, the assumption was wrong,\(\neg P(a)\)is true.

Hence, the premises\(\forall x(P(x) \to Q(x))\) and\(\neg Q(a)\) for a particular element a in the domain, imply\(\neg P(a)\).