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Proof step 2 from proof mu-is-IS (previous step)

⊦ μ(f)(x) ↔ IS(λ S. f(S) ⊆ S)(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ IS(SS)(x) ↔ (∀ S. SS(S) → S(x))
Proof step 4:
⊦ μ(f)(x) ↔ (∀ S. f(S) ⊆ S → S(x))

Rule argument

spec: inst: name: intersection-def seq: ⊦ IS(SS)(x) ↔ (∀ S. SS(S) → S(x)) subst: [x, SS] → [x, λ S. f(S) ⊆ S] flags: rewrite, current

Additional information

Used terms: x, λ S. f(S) ⊆ S