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I just need a little help finishing off this proof, not sure where to go from here with introducing (∀x)

(∀x) F(x) ⇒ G(x), (∃x) F(x) ⊢ (∀x) G(x)

1 (1) (∀x) F(x) ⇒ G(x) A

2 (2) (∃x) F(x) A

3 (3) F(a) A

1 (4) F(a) ⇒ G(a) 1 ∀ E

1, 3 (5) G(a) 3, 4 MP

Thanks

feeblefeeble
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  • I'm pretty sure you can only generalize to (∃x) G(x), not (∀x) G(x). You got line 3 by existential instantiation, which means that your "a" is some specific value (one for which it holds that F(a)), and not a free variable. – Matt Dickau May 17 '16 at 14:37
  • You cannot prove it, because it is not valid. – Mauro ALLEGRANZA May 17 '16 at 15:05

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