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Proof step 5 from proof sub-one (previous step)

∀ a. a ∈ s → a ∈ s0, ∃ a. s0 = ∅ ++ a ⊦ # s = 0, s = ∅, s = s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a0. a0 ∈ s → a0 = a ⊦ # s = 0, s = ∅ ++ a, s = ∅

Rule argument

Additional information

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