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

∀ a0. a0 ∈ ∅ ++ a ++ b ↔ a0 ∈ ∅ ++ c ++ b ⊦ a = c

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
b ∈ ∅ ++ c, ∀ a0. a0 ∈ ∅ ++ a ++ b ↔ a0 ∈ ∅ ++ c ++ b ⊦ a = c
Proof step 5:
∀ a0. a0 ∈ ∅ ++ a ++ b ↔ a0 ∈ ∅ ++ c ++ b ⊦ a = c, b ∈ ∅ ++ c

Rule argument

fma: b ∈ ∅ ++ c

Additional information

Used formulas: b ∈ ∅ ++ c