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Proof step 3 from proof eqtwo-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
b ∈ ∅ ++ c, ∀ a0. a0 ∈ ∅ ++ a ++ b ↔ a0 ∈ ∅ ++ c ⊦ a = c

Rule argument

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;