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Proof step 24 from proof nu-is-UN (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
S ⊆ f(S), S(x) ⊦ compl(f)(λ x0. ¬ S(x0)) ⊆ (λ x0. ¬ S(x0))

Rule argument

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