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Proof step 28 from proof nu-is-UN (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ compl(f)(S)(x) ↔ ¬ f(λ x0. ¬ S(x0))(x)
Proof step 30:
S(x0), ¬ f(λ x0. ¬ ¬ S(x0))(x0), S ⊆ f(S), S(x) ⊦

Rule argument

spec: inst: name: compl-def seq: ⊦ compl(f)(S)(x) ↔ ¬ f(λ x0. ¬ S(x0))(x) subst: [x, S, f] → [x0, λ x0. ¬ S(x0), f] flags: rewrite, current

Additional information

Used terms: x0, λ x0. ¬ S(x0), f