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Prove the following formula in Fitch format:

∃x(∀y(P(y)→y=x)∧P(a)) |= ∀x∀y(¬(x=y)→(¬P(x)∨¬P(y)))

I tried to use universal introduction as my main rule but didn't know how to proceed

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    I would love to see what the proof looks like! You should show yer work, how much progress you've made if your thread is to get any hits.
    – Agent Smith
    Jun 5 at 3:22
  • As a sketch, I suggest: assume P(b) ∧ P(c), where b and c are 'boxed constants'. Use existential instantiation on your given premise and prove b=c, then discharge the assumption to get (P(b) ∧ P(c)) → b=c. Then contrapose to get ¬(b=c) → ¬(P(b) ∧ P(c)) and hence by de Morgan, ¬(b=c) → (¬P(b) ∨ ¬P(c)). Then by two applications of universal generalisation get ∀x∀y(¬(x=y) → (¬P(x) ∨ ¬P(y))).
    – Bumble
    Jun 5 at 18:04

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