Use the rules of inference to show that if Vx (Ax) V Qx) and Vx ((-PX)...

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Use the rules of inference to show that if Vx (Ax) V Qx) and Vx ((-PX) A QX) - R(X) are true, then V xt-Rx) - Px) is also true, where the domains of all quantifiers are the same. Construct your argument by rearranging the following building blocks. 1. We will show that if the premises are true, then (-R(a) – P(a)) for every a. 2. Suppose ¬R(a) is true for some a. 3. For such an a, universal modus tollens applied to the second premise gives us (-P(a)^Q(a). Drag the text blocks below into their correct order. X(-R(x) P(x)). (-P(a) a Q(a). By universal generalization, we get, P(a) v P(a). P(a) v Q(a). Applying the rules of De Morgan's law on By resolution, we conclude We have therefore shown R(a) – P(a) for every a. By universal instantiation on Vx (P(x) v Q(x)). we conclude This is logically equivalent to P(a). We get, P(a) v ¬Q(a). Use the rules of inference to show that if Vx (Ax) V Qx) and Vx ((-PX) A QX) - R(X) are true, then V xt-Rx) - Px) is also true, where the domains of all quantifiers are the same. Construct your argument by rearranging the following building blocks. 1. We will show that if the premises are true, then (-R(a) – P(a)) for every a. 2. Suppose ¬R(a) is true for some a. 3. For such an a, universal modus tollens applied to the second premise gives us (-P(a)^Q(a). Drag the text blocks below into their correct order. X(-R(x) P(x)). (-P(a) a Q(a). By universal generalization, we get, P(a) v P(a). P(a) v Q(a). Applying the rules of De Morgan's law on By resolution, we conclude We have therefore shown R(a) – P(a) for every a. By universal instantiation on Vx (P(x) v Q(x)). we conclude This is logically equivalent to P(a). We get, P(a) v ¬Q(a).