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Proof step 11 from proof nu-is-UN (previous step)

compl(f)(S) ⊆ S ⊦ S(x), ∃ S. S ⊆ f(S) ∧ S(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 13:
∀ x. compl(f)(S)(x) → S(x) ⊦ S(x), ∃ S. S ⊆ f(S) ∧ S(x)

Rule argument

spec: inst: name: subseteq-def seq: ⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x)) subst: [S0, S] → [S, compl(f)(S)] flags: rewrite, current

Additional information

Used terms: S, compl(f)(S)