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Proof step 7 from proof compl-monotone (previous step)

S0 ⊆ S, ∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0) ⊦ compl(f)(S0) ⊆ compl(f)(S)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 9:
S0 ⊆ S, ∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0) ⊦ ∀ x. compl(f)(S0)(x) → compl(f)(S)(x)

Rule argument

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

Additional information

Used terms: compl(f)(S), compl(f)(S0)