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Proof step 2 from proof int-inserte (previous step)

a ∈ s, a0 ∈ s, ∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ++ a ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic Used terms: a