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Proof step 7 from proof size-two-06 (previous step)

#(s0 ++ a0) < 2, a ∈ s0, ∀ a, b. # s0 < 2 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ a = b

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
#(s0 ++ a0) < 2, a ∈ s0, ∀ a, b. # s0 < 2 ∧ a ∈ s0 ∧ b ∈ s0 → a = b, b = a0 ∨ b ∈ s0 ⊦ b = a

Rule argument

fma: b = a

Additional information

Used formulas: b = a