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Proof step 4 from proof inex (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-putsame: ⊦ a ∈ st → (st -- a)[a, st[a]] = st;%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;