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Proof step 6 from proof nu-is-UN (previous step)

⊦ ¬ μ(compl(f))(x) ↔ (∃ S. S ⊆ f(S) ∧ S(x))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ μ(f)(x) ↔ (∀ S. f(S) ⊆ S → S(x))
Proof step 8:
⊦ ¬(∀ S. compl(f)(S) ⊆ S → S(x)) ↔ (∃ S. S ⊆ f(S) ∧ S(x))

Rule argument

spec: inst: name: least-prefixpoint-def seq: ⊦ μ(f)(x) ↔ (∀ S. f(S) ⊆ S → S(x)) subst: [x, f] → [x, compl(f)] flags: rewrite, current

Additional information

Used terms: x, compl(f)