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Proof step 9 from proof nu-is-UN (previous step)

⊦ ¬(∀ 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 10:
¬(∀ S. compl(f)(S) ⊆ S → S(x)) ⊦ ∃ S. S ⊆ f(S) ∧ S(x)
Proof step 22:
∃ S. S ⊆ f(S) ∧ S(x) ⊦ ¬(∀ S. compl(f)(S) ⊆ S → S(x))

Rule argument

pos: right 1

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