1. A v (B -> C) 2. B / A v C 3.
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Originally Posted by ambergarrett955 1. A v (B -> C) 2. B 3./ A v C Do by indirect argument. Assume ~(AvC) then ~A ^ ~C. ~A by simplification. (B->C) by disjunctive syllogism. ~C by simplification. ~B by modus tollens. That is a contradiction.
I'm trying to get A v C by itselfs, though. So, is that still the answer? Thank you
Originally Posted by ambergarrett955 I'm trying to get A v C by itselfs, though. So, is that still the answer? The answer is yes only if you know how to use indirect argument. If you do not have that tool, then I do not how to help you with this.
In class we always came out with the answer that was behind this /xx as our last #. I'm just confused. This stuff is so confusing to me. I'm sorry.
Hello, Amber! I vaguely recall a theorem . . . the Law of Detachment (?) . . The first line is: . So the argument becomes: . By the Law of Detachment, the conclusion is: .
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