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Proof step 5 from proof del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ s0, a0 ∈ s0, ∀ a. a ∈ s0 → (∃ s1. s0 = s1 ++ a ∧ ¬ a ∈ s1) ⊦ ∃ s. s0 = s ++ a ∧ ¬ a ∈ s

Rule argument

Additional information

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