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Proof step 5 from proof rem-eq (previous step)

a ∈ st, ∀ a0. a0 ∈ st -- a ↔ a0 ∈ st ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a

Additional information

Used simplifier rules: notin-del-notin: ⊦ ¬ a ∈ st → st -- a = st;%Del-notin: ⊦ ¬ a ∈ st -- a; Used terms: a