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Proof step 18 from proof size-two (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 19:
s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, a ∈ s0, b ∈ ∅ ++ a0

Rule argument

fma: b ∈ ∅ ++ a0

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%eq-02: ⊦ ¬ a ∈ s → (s ++ a = ∅ ++ b ↔ a = b ∧ s = ∅);%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b; Used formulas: b ∈ ∅ ++ a0