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Proof step 5 from proof prefix-ex (previous step)

a ' + x0 ⊑ y + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ a ' + x0 ⊑ y, ∃ x. a ' + x0 = y + x ∧ x ⊑ z

User applied rule cut formula

And obtained the following new goals

Proof step 6:
a ' + x0 ⊑ y + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ y = [], a ' + x0 ⊑ y, ∃ x. a ' + x0 = y + x ∧ x ⊑ z

Rule argument

fma: y = []

Additional information

Used simplifier rules: prefix-empty: ⊦ x ⊑ [] ↔ x = [];% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s) Used formulas: y = []