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Proof step 8 from proof del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ s0, ∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1) ⊦ a0 ∈ s0, ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

Rule argument

Additional information

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