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Proof step 3 from proof indef (previous step)

a ∈ st ⊦ ∀ a0. (a0 ∈ st ↔ a0 ∈ st[a, st[a]]) ∧ st[a0] = st[a, st[a]][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ st ⊦ (a0 ∈ st → a0 ∈ st[a, st[a]]) ∧ st[a0] = st[a, st[a]][a0]

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;