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Proof step 1 from proof dom-ins (conclusion)

⊦ dom st[a, d] = dom st ++ a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)

Rule argument

spec: set-basic inst: name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [dom st ++ a, dom st[a, d]] flags: rewrite

Additional information

Used simplifier rules: Dom: ⊦ a ∈ dom st ↔ a ∈ st;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st); from specification store%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st; from specification store Used terms: dom st ++ a, dom st[a, d]