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Proof step 15 from proof size-two (previous step)

∃ a0, b. a0 ≠ b ∧ s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, a ∈ s0

Rule argument

Additional information