(a) (∀x(Fx ∨ Gx))(a,b)
Phil 160 - HW#9 1. Compute the following truth-functional expansions: (a) (b) (c) (d)
(∀x(F x ∨ Gx))(a,b) (∀xF x ∨ ∀xGx)(a,b) (∀x∃yRxy)(a,b) (∃x∀yRxy)(a,b)
2. Construct possible worlds to establish the invalidity of. . . (a) ∀x(F x → Gx), Ga ∴ F a (b) ∀x(F x ∨ Gx) ∴ ∀xF x ∨ ∀xGx (c) ∀x∃yRxy ∴ ∃x∀yRxy 3. Translate the following sentences into formulas of quantificational logic. Then, and establish their invalidity by constructing a possible world in which the premises are true and the conclusion is false. (You won’t need to make a world with more than two objects in its domain.) (a) All F s are Gs, but a not an F . Therefore, a is not a G. (b) All F s are Gs. But some Gs are not Hs. So, some F s are not Hs. 4. (Bonus!) Consider the argument ∀x∃yRxy, ∀x∀y∀z(Rxy ∧ Ryz → Rxz) ∴ ∃xRxx. Is it valid? Or can you find a countermodel?
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(∀x(F x ∨ Gx))(a,b) (∀xF x ∨ ∀xGx)(a,b) (∀x∃yRxy)(a,b) (∃x∀yRxy)(a,b)
2. Construct possible worlds to establish the invalidity of. . . (a) ∀x(F x → Gx), Ga ∴ F a (b) ∀x(F x ∨ Gx) ∴ ∀xF x ∨ ∀xGx (c) ∀x∃yRxy ∴ ∃x∀yRxy 3. Translate the following sentences into formulas of quantificational logic. Then, and establish their invalidity by constructing a possible world in which the premises are true and the conclusion is false. (You won’t need to make a world with more than two objects in its domain.) (a) All F s are Gs, but a not an F . Therefore, a is not a G. (b) All F s are Gs. But some Gs are not Hs. So, some F s are not Hs. 4. (Bonus!) Consider the argument ∀x∃yRxy, ∀x∀y∀z(Rxy ∧ Ryz → Rxz) ∴ ∃xRxx. Is it valid? Or can you find a countermodel?
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