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Proof step 3 from proof eqtwo-01 (previous step)

∀ a0. a0 ≠ b → (a0 = a ↔ a0 = c) ⊦ a = c, b ∈ ∅ ++ c

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a0. a0 ≠ b → (a0 = a ↔ a0 = c) ⊦ a = c, b = c

Rule argument

Additional information

Used simplifier rules: in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;