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Proof step 15 from proof nu-is-UN (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 17:
∀ x. compl(f)(S)(x) → S(x) ⊦ S(x), ∀ x. ¬ S(x) → f(λ x0. ¬ S(x0))(x)

Rule argument

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

Additional information

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