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Proof step 3 from proof del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1), a = a0 ∨ a ∈ s0 ⊦ ∃ s1. s0 ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

Rule argument

Additional information

Used simplifier rules: rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);