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Proof step 3 from proof eq-bothsubset (previous step)

⊦ S = S0 ↔ (∀ x. S(x) → S0(x)) ∧ S0 ⊆ S

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 5:
⊦ S = S0 ↔ (∀ x. S(x) → S0(x)) ∧ (∀ x. S0(x) → S(x))

Rule argument

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

Additional information

Used terms: S, S0