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Proof step 3 from proof two (previous step)

a = b ∨ a = a ↔ a = c ∨ a = a0, ∀ a1. a1 = b ∨ a1 = a ↔ a1 = c ∨ a1 = a0 ⊦ a = a0, a = c ∧ b = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information