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Proof step 5 from proof del-three (previous step)

#(s -- a) - 1 = 0, 2 < # s, b ∈ s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b = a

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
#(s -- a) - 1 = 0, 2 < # s, a0 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b = a
Proof step 10:
#(s -- a) - 1 = 0, 2 < # s, a0 ∈ s, b ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b = a, a ∈ s

Rule argument

fma: a ∈ s

Additional information

Used formulas: a ∈ s