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Proof step 5 from proof eqtwo-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a0. a0 ∈ ∅ ++ a ++ b ↔ a0 ∈ ∅ ++ c ++ b ⊦ a = c, b = c

Rule argument

Additional information

Used simplifier rules: add: ⊦ a ∈ s ++ b ↔ ¬(a ≠ b ∧ ¬ a ∈ s);%In-empty: ⊦ ¬ a ∈ ∅;