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If I have ∃x(Fx implies P) then I can clearly instantiate to (Fy implies P) by existential instantiation. What if I have instead ((∃x Fx) implies P) then can I existentially instantiate to (Fy implies P)?

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    NO; if you are working with e.g.Natural Deduction or Fitch, you must first "unpack" the formula ((∃x Fx) → P) according to the main "logical opearor", that in this case is . – Mauro ALLEGRANZA Nov 2 '17 at 19:05
  • Maybe you have to start assuming Fx, derive (∃x Fx) by ∃-intro and finally use →-elim to derive P from ((∃x Fx) → P). – Mauro ALLEGRANZA Nov 2 '17 at 19:06
  • I think this actually requires the Axiom of Choice. Most automatic prover programs assume it implicitly. You may have an infinity of xs meeting condition F(x), but that doesn't mean you can choose a specific one! – barrycarter Nov 3 '17 at 17:00
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Yes, every y instantiates (Fy implies P):

Choose any y at all.

We are given that ((Exists x: Fx) implies P).
Then either P is true or (Exists x: Fx) is false.

Assume P is true.
Then (Fy implies P), because anything implies true.

Otherwise, (Exists x: Fx) is false.
Therefore Fy is false for our y.
Then (Fy implies P), because false implies anything.
  • I don't understand "Then (Fy implies P), because anything implies true." While it is true that ((False or True) implies True) is always True, I don't see how it is relevant, perhaps I'm missing your point? – Clclstdnt Nov 2 '17 at 19:18
  • What would make it true but not relevant, in your interpretation of what is going on? This proof fails for logics that involve a notion of relevance: proof-theory, constructive reasoning, Intuitionism, etc. But normal classical logic based on models just doesn't have such a concept AFAIK. – user9166 Nov 2 '17 at 20:53
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{1}     1.  Ǝx[Fx] → P             Prem.
{2}     2.  Ǝy[Fy]                 Assum.
{3}     3.  Fa                     Assum. TD(a)
{3}     4.  Ǝx[Fx]                 3 EI
{1,3}   5.  P                      1,4 MP
{1,2}   6.  P                      2,3,5 EE
{1}     7.  Ǝy[Fy] → P             2,6 CP

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