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Proof step 1 from proof eqtwo-01 (conclusion)

⊦ ∅ ++ a ++ b = ∅ ++ c ++ b ↔ a = c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a0. a0 ≠ b → (a0 = a ↔ a0 = c) ⊦ a = c

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;