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Proof step 7 from proof sub-one (previous step)

∀ a0. a0 ∈ s → a0 = a ⊦ # s = 0, s = ∅, s = ∅ ++ a, a ∈ ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ a0. a0 ∈ s → a0 = a ⊦ # s = 0, s = ∅ ++ a, s = ∅

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅;