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Proof step 14 from proof nu-is-UN (previous step)

∀ x. compl(f)(S)(x) → S(x) ⊦ S(x), (λ x0. ¬ S(x0)) ⊆ f(λ x0. ¬ S(x0)) ∧ ¬ S(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
∀ x. compl(f)(S)(x) → S(x) ⊦ S(x), (λ x0. ¬ S(x0)) ⊆ f(λ x0. ¬ S(x0))

Rule argument

Additional information