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Proof step 12 from proof del-three (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 13:
a ∈ s, #(s -- a) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Rule argument

fma: a ∈ s

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;%f: ⊦ m < n → n ≠ 0; from specification nat-basic1 Used formulas: a ∈ s