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Proof step 5 from proof prefix-split-stronger (previous step)

∀ a, x0, z. a ' + x0 ⊑ y0 + z ∧ (∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z)) → a ' + x0 ⊑ y0 ∨ (∃ x. a ' + x0 = y0 + x ∧ x ⊑ z), a ' + x0 ⊑ (a0 + y0) + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ a ' + x0 ⊑ a0 + y0, ∃ x. a ' + x0 = (a0 + y0) + x ∧ x ⊑ z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
x0 ⊑ y0 + z, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z), ∀ a, x0, z. a ' + x0 ⊑ y0 + z ∧ (∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z)) → a ' + x0 ⊑ y0 ∨ (∃ x. a ' + x0 = y0 + x ∧ x ⊑ z) ⊦ x0 ⊑ y0, ∃ x. x0 = y0 + x ∧ x ⊑ z

Rule argument

Additional information

Used simplifier rules: prefix-rec-03: ⊦ a ' + x ⊑ b ' + y ↔ a = b ∧ x ⊑ y;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list