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Proof step 6 from proof del (previous step)

a ∈ s0, a0 ∈ s0, ∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1) ⊦ ∃ s. s0 = s ++ a ∧ ¬ a ∈ s

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);%in-03: ⊦ ¬ a ∈ s → (a ∈ s ++ b ↔ a = b);%eq-01: ⊦ ¬ a ∈ s ∧ ¬ a ∈ s0 → (s ++ a = s0 ++ a ↔ s = s0); Used terms: a