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Proof step 7 from proof eqtwo-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
c ∈ ∅ ++ a ++ b, c ∈ ∅ ++ c ++ b, ∀ 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);