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Proof step 6 from proof lastoc (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 7:
a ∈ x0, ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ z), a = a0 ∨ a ∈ x0, a ∈ x0 → x0 = y + a ' + z ∧ ¬ a ∈ z ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ z
Proof step 9:
∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ z), a = a0 ∨ a ∈ x0, a ∈ x0 → x0 = y + a ' + z ∧ ¬ a ∈ z ⊦ a ∈ x0, ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ z

Rule argument

fma: a ∈ x0

Additional information

Used formulas: a ∈ x0