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Proof step 2 from proof two (previous step)

⊦ ∅ ++ a ++ b = ∅ ++ a0 ++ c ↔ a = c ∧ b = a0, a = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a1. a1 = b ∨ a1 = a ↔ a1 = c ∨ a1 = a0 ⊦ a = a0, a = c ∧ b = a0

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%eq-04: ⊦ s ++ b ++ a = s ++ a ++ b ↔ true;