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Proof step 6 from proof indef (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: a

Additional information

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