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Proof step 1 from proof in-05 (conclusion)

⊦ s ++ a = s0 ++ b ∧ a ≠ b → a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a0. a0 = a ∨ a0 ∈ s ↔ a0 = b ∨ a0 ∈ s0 ⊦ a = b, a ∈ s0

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);