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Proof step 5 from proof indef (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information