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Proof step 4 from proof in-def (previous step)

∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a = a0, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z
Proof step 7:
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z), a = a0 ∨ a ∈ x0 ⊦ a = a0, ∃ y, z. a0 ' + x0 = y + a ' + z

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0