Question

Justify the rule of universal transitivity, which states that if\(\forall {\rm{x(P(x)}} \to {\rm{Q(x))}}\) and \(\forall {\rm{x(Q(x)}} \to R{\rm{(x))}}\) are true, then\(\forall {\rm{x(P(x)}} \to R{\rm{(x))}}\) is true, where the domains of all quantifiers are the same.

Short answer

If\(\forall x(P(x) \to Q(x))\) and \(\forall x(Q(x) \to R(x))\) are true, this implies\(\forall x(P(x) \to R(x))\) is also true which justifies the Universal Transitivity.

Step-by-step solution

Rules of Inference

(a) Hypothetical Syllogism: \(((p \to q) \wedge (q \to r)) \to (p \to r)\)

(b) 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\(\forall xP(x)\).

(c) Universal generalization: It is a rule of inference that states that \(\forall xP(x)\) is true, given the premise that P(a) is true for all elements a in the domain.

Proof

Given: Premises\(\forall x(P(x) \to Q(x))\)and\(\forall x(Q(x) \to R(x))\).

To show:\(\forall x(P(x) \to R(x))\)

Proof: Valid argument that the premises lead to the desired conclusion can be given as-

Step Reason

1. \(\forall {\rm{x(P(x)}} \to {\rm{Q(x))}}\) Premise
2. \(P(a) \to Q(a)\) Universal Instantiation from (1)
3. \(\forall {\rm{x(Q(x)}} \to R{\rm{(x))}}\) Premise
4. \(Q(a) \to R(a)\) Universal Instantiation from (3)
5. \(P(a) \to R(a)\) Hypothetical Syllogism using (2) and (4)
6. \(\forall {\rm{x(P(x)}} \to R{\rm{(x))}}\) Universal Generalization from (5)

HenceUniversal Transitivityis justified since premises\(\forall x(P(x) \to Q(x))\) and \(\forall x(Q(x) \to R(x))\) imply\(\forall x(P(x) \to R(x))\) is also true.