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Proof step 4 from proof size-two-04 (previous step)

∃ a0. a0 ∈ s ∧ ¬ a0 ∈ ∅ ++ a, a ∈ s ⊦ # s = 0, # s = 1, ∃ b. b ∈ s ∧ a ≠ b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ s, a ∈ s ⊦ # s = 0, # s = 1, a0 = a, ∃ b. b ∈ s ∧ a ≠ b

Rule argument

Additional information

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