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Proof step 4 from proof subset-trans (previous step)

S0 ⊆ S1, ∀ x. S(x) → S0(x) ⊦ S ⊆ S1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 6:
∀ x. S0(x) → S1(x), ∀ x. S(x) → S0(x) ⊦ S ⊆ S1

Rule argument

spec: inst: name: subseteq-def seq: ⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x)) subst: [S0, S] → [S1, S0] flags: rewrite, current

Additional information

Used terms: S1, S0