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Proof step 1 from proof indef (conclusion)

⊦ a ∈ st ↔ st = st[a, st[a]]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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