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Proof step 2 from proof eqtwo-01 (previous step)

∀ a0. a0 ≠ b → (a0 = a ↔ a0 = c) ⊦ a = c

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
∀ a0. a0 ≠ b → (a0 = a ↔ a0 = c) ⊦ a = c, b ∈ ∅ ++ c

Rule argument

fma: b ∈ ∅ ++ c

Additional information

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