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Proof step 1 from proof del (conclusion)

a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);