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Drag the text blocks below into their correct order. By universal generalization, we get, P(a) v Q(a). x(x)→ P(x)). Applying the rules of De Morgan's law on (-P(a) ^ Q(a)). A This is logically equivalent to P(a). By universal instantiation on Vx (P(x) v Q(x)). we conclude We have therefore shown R(a) → P(a) for every a. We get, P(a) v ¬Q(a). P(a) v P(a). By resolution, we conclude

Drag the text blocks below into their correct order. By universal generalization, we get, P(a) v Q(a). x(x)→ P(x)). Applying the rules of De Morgan's law on (-P(a) ^ Q(a)). A This is logically equivalent to P(a). By universal instantiation on Vx (P(x) v Q(x)). we conclude We have therefore shown R(a) → P(a) for every a. We get, P(a) v ¬Q(a). P(a) v P(a). By resolution, we conclude

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Use the rules of inference to show that if Væ(P(x) VQ(x)) and Vx((-P(x) ^ Q(x))→ R(x)) are true, then Vr(-R(x) P(x)) 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. 00000000 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)). By universal generalization, we get, P(a) v Q(a). Vx (R(x)→ P(x)). Applying the rules of De Morgan's law on (~P(a) ^ Q(a)). This is logically equivalent to P(a). 0000 Drag the text blocks below into their correct order. By universal instantiation on Vx (P(x) v Q(x)). we conclude We have therefore shown R(a) P(a) for every a. We get, P(a) v ¬Q(a). P(a) v P(a). By resolution, we conclude
Transcribed Image Text:Use the rules of inference to show that if Væ(P(x) VQ(x)) and Vx((-P(x) ^ Q(x))→ R(x)) are true, then Vr(-R(x) P(x)) 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. 00000000 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)). By universal generalization, we get, P(a) v Q(a). Vx (R(x)→ P(x)). Applying the rules of De Morgan's law on (~P(a) ^ Q(a)). This is logically equivalent to P(a). 0000 Drag the text blocks below into their correct order. By universal instantiation on Vx (P(x) v Q(x)). we conclude We have therefore shown R(a) P(a) for every a. We get, P(a) v ¬Q(a). P(a) v P(a). By resolution, we conclude
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