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Proof step 24 from proof size-two (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 25:
s0 ++ a = ∅ ++ a0 ++ b ⊦ 1 = # s0, a0 = b, s0 = ∅ ++ b, a ∈ s0, b ∈ ∅, ∃ a. s0 = ∅ ++ a

Rule argument

fma: b ∈ ∅

Additional information

Used simplifier rules: eq: ⊦ a ∈ s → s ++ a = s;%In-empty: ⊦ ¬ a ∈ ∅; Used formulas: b ∈ ∅