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Proof step 25 from proof size-two (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, a ∈ s0, ∃ a. s0 = ∅ ++ a

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅;