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Proof step 22 from proof nu-is-UN (
previous step
)
∃ S. S ⊆ f(S) ∧ S(x) ⊦ ¬(∀ S. compl(f)(S) ⊆ S → S(x))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 23
:
S ⊆ f(S), S(x), ∀ S. compl(f)(S) ⊆ S → S(x) ⊦
Rule argument
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