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Proof step 25 from proof nu-is-UN (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 27:
S ⊆ f(S), S(x) ⊦ ∀ x. compl(f)(λ x0. ¬ S(x0))(x) → ¬ S(x)

Rule argument

spec: inst: name: subseteq-def seq: ⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x)) subst: [S0, S] → [λ x0. ¬ S(x0), compl(f)(λ x0. ¬ S(x0))] flags: rewrite, current

Additional information

Used terms: λ x0. ¬ S(x0), compl(f)(λ x0. ¬ S(x0))