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Proof step 10 from proof sub-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ a0. a0 ∈ s → a0 = a ⊦ # s = 0, a0 ∈ s ↔ a0 = a, s = ∅

Rule argument

Additional information

Used simplifier rules: in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;