Question

Use rules of inference to show that if \(\forall x(P(x) \to ((Q(x) \wedge S(x)))\) and\(\forall x(P(x) \wedge R(x))\)are true, then\(\forall x(R(x) \wedge S(x))\)is true.

Short answer

\(\forall x(P(x) \to ((Q(x) \wedge S(x)))\)and\(\forall x(P(x) \wedge R(x))\)imply that \(\forall x(R(x) \wedge S(x))\)is true.

Step-by-step solution

Rules of Inference needed in the proof

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

(b) Universal generalization: This rule states that ∀xP(x) is true, given the premise that P(a) is true for all elements a in the domain.

(c) Modus Ponens: \(((P(a) \wedge (P(a) \to Q(a)) \to Q(a)\) where a belongs to the domain.

(d) Simplification: \((P(a) \wedge Q(a)) \to P(a)\) where a belongs to the domain.

(e) Conjunction: \(((P(a)) \wedge (Q(a))) \to (P(a) \wedge Q(a))\) where a belongs to the domain.

Proof

Given premises are \(\forall x(P(x) \to ((Q(x) \wedge S(x)))\)and \(\forall x(P(x) \wedge R(x))\)

UsingUniversal instantiation, it can be concluded that

\((P(a) \to ((Q(a) \wedge S(a)))\) (1)

and\((P(a) \wedge R(a))\) (2)

UsingSimplificationRule on (2)

\((P(a) \wedge R(a)) \to P(a)\) (3)

or\((P(a) \wedge R(a)) \to R(a)\) (by commutative property) (4)

ApplyingModus Ponenson (1) and conclusion of (3)

\((P(a) \wedge (P(a) \to (Q(a) \wedge S(a)))) \to (Q(a) \wedge S(a))\) (5)

Using Simplification rule on (5)

\((Q(a) \wedge S(a)) \to S(a)\) (6)

Applying Conjunction Rule on conclusions of (4) and (6)

\(((R(a)) \wedge (S(a))) \to (R(a) \wedge S(a))\)

Thus, according to Universal generalization

\(\forall x(R(x) \wedge S(x))\)

Hence \(\forall x(R(x) \wedge S(x))\) is true.