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Proof step 11 from proof prefix-ex (previous step)

x0 ⊑ y0 + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ x0 ⊑ y0, ∃ x. x0 = y0 + x ∧ x ⊑ z

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: y0, z abort: F discard: T computed: T

Additional information

Used simplifier rules: postifx-app-03: ⊦ x + y ⊑ x + z ↔ y ⊑ z;%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + y = x + z ↔ y = z; (s) Used terms: y0, z