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Proof step 1 from proof ins (conclusion)

⊦ st[a] = d ∧ a ∈ st → st[a, d] = st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]);%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;