(a) (1) (x)[Ax → (∃y)(Kxy)] ∴ (C) (∃y)(x)[Ax → Kxy] (b) (1) (x)[(Px & Qx) → Rx] (2) (∃x)[Qx & ~Rx] (3) (∃z)[Pz & ~Rz] ∴ (C) (∃y)[~Py & ~Qy] (c) (1) (∃x)[Ax & Bx] (2) (x)[~Bx v ~Cx] ∴ (C) (x)[~Ax v ~Cx] (d) (1) (x)[Fx → Gx] → (y)[Hy → Jy] ∴ (C) (∃x)[Fx & Gx] → (y)[Hy → Jy]
Added by Michelle T.
Step 1
Let's use a domain of discourse with two objects, a and b. Let Aa be true and Abb be false. Let Kaa and Kab be true. Then, the premise (1) is true because Ax is true for both a and b, and there exists a y (in this case, y = a) such that Kxy is true. However, the Show more…
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