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Proof step 23 from proof nu-is-UN (previous step)

S ⊆ f(S), S(x), ∀ S. compl(f)(S) ⊆ S → S(x) ⊦

User applied rule all left

And obtained the following new goals

Proof step 24:
S ⊆ f(S), S(x), compl(f)(λ x0. ¬ S(x0)) ⊆ (λ x0. ¬ S(x0)) → ¬ S(x) ⊦

Rule argument

pos: left 3 terms: λ x0. ¬ S(x0) abort: F discard: T computed: F

Additional information

Used terms: λ x0. ¬ S(x0)