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Proof step 10 from proof del (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ a1. a1 = a ∨ a1 ∈ s → (∃ s1. s ++ a = s1 ++ a1 ∧ ¬ a1 ∈ s1) ⊦ a0 = a, a ∈ s, a0 ∈ s, ∃ s1. s ++ a ++ a0 = s1 ++ a ∧ ¬ a ∈ s1

Rule argument

Additional information

Used simplifier rules: in-03: ⊦ ¬ a ∈ s → (a ∈ s ++ b ↔ a = b);%add: ⊦ ¬ a ∈ s ++ b ↔ a ≠ b ∧ ¬ a ∈ s;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);