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Proof step 2 from proof del-three (previous step)

#(s -- a -- b) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
b ∈ s -- a, #(s -- a -- b) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦
Proof step 11:
#(s -- a -- b) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b ∈ s -- a

Rule argument

fma: b ∈ s -- a

Additional information

Used formulas: b ∈ s -- a