Question

Use rules of inference to show that if the premises \(\forall xP\left( x \right) \to Q\left( x \right)\), \(\forall x\left( {Q\left( x \right) \to R\left( x \right)} \right)\), and \(\neg R\left( a \right)\), where a is in the domain, are true, then the conclusion \(\neg P\left( a \right)\) is true.

Short answer

The \(\neg P\left( a \right)\) is true.

Step-by-step solution

Rule of Inference

Modus tollens is \(\frac{{p\mathop \to \limits^{\neg q} q}}{{\neg p}}\).

Universal instantiation is \(\frac{{\forall xP\left( x \right)}}{{P\left( c \right)}}\).

Determine whether the conclusion \(\neg P\left( a \right)\) is true

Assume that the premises are true.

	Step	Reason
1.	\(\forall x\left( {P\left( x \right) \to Q\left( x \right)} \right)\)	Premise
2.	\(\forall x\left( {Q\left( x \right) \to R\left( x \right)} \right)\)	Premise
3.	\(\neg R\left( a \right)\)	Premise
4.	\(P\left( a \right) \to Q\left( a \right)\)	Universal instantiation from (1)
5.	\(Q\left( a \right) \to R\left( a \right)\)	Universal instantiation from (2)
6.	\(\neg Q\left( a \right)\)	Modus tollens from (3) and (5)
7.	\(\neg P\left( a \right)\)	Modus tollens from (4) and (6)

Therefore, it is shown that \(\neg P\left( a \right)\) is true when the premises are true.