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Proof step 2 from proof indef (previous step)

a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 3:
a ∈ s, a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0

Rule argument

fma: a ∈ s

Additional information

Used formulas: a ∈ s