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Proof step 1 from proof insins (conclusion)

⊦ st[a, d0][a, d] = st[a, d]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (a0 ∈ st[a, d0][a, d] ↔ a0 ∈ st[a, d]) ∧ st[a, d0][a, d][a0] = st[a, d][a0]

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]);