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Proof step 22 from proof size-two (previous step)

s0 ++ a = ∅ ++ a0 ++ b, # s0 = 1 ↔ (∃ a. s0 = ∅ ++ a) ⊦ 1 = # s0, a0 = b, a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

Additional information