Demonstrate that the following arg are valid by means of a proof. You ? any of the three methods proof. (Ex used as the existential quantifier. (x)[(Ax v Bx) \to Cx] / (Ex)Bx // (Ex)C 1. Show (Ex)Cx (x)(Gx \to Hx) / (Ex)(~Fx \cdot Gx) // (Ex) 1. Show (Ex)(~Fx \cdot Hx) ~(Ex)(Rx v Sx) v (x)Tx / (Ex)~Tx // ~ 1. Show ~(Ex) Sx
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Show (Ex)Cx -> (x)(Gx -> Hx) Proof by contradiction: Assume (Ex)Cx is true, but (x)(Gx -> Hx) is false. This means there exists an object x such that Cx is true, but for all objects x, Gx -> Hx is false. Let's consider this object x. Since Cx is true, we can Show more…
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