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

monotone(compl(f)), monotone(f) ⊦ ¬(∀ S. compl(f)(S) ⊆ S → S(x)) ↔ (∃ S. S ⊆ f(S) ∧ S(x))

User applied rule case distinction

And obtained the following new goals

Proof step 15:
monotone(f), monotone(compl(f)) ⊦ ∀ S. compl(f)(S) ⊆ S → S(x), ∃ S. S ⊆ f(S) ∧ S(x)
Proof step 27:
monotone(f), monotone(compl(f)), ∀ S. compl(f)(S) ⊆ S → S(x), ∃ S. S ⊆ f(S) ∧ S(x) ⊦

Rule argument

pos: right 1

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