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

x = a0 + y0 ∧ x ≠ [], x ⊑ y, y ⊑ a ' + z0, y0 ⊑ y.rest ∧ y.rest ⊑ z0 → y0 ⊑ z0, ∀ x, y. x ⊑ y ∧ y ⊑ z0 → x ⊑ z0 ⊦ x ⊑ a ' + z0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
y ⊑ a ' + z0, a0 ' + y0 ⊑ y, y0 ⊑ y.rest ∧ y.rest ⊑ z0 → y0 ⊑ z0, ∀ x, y. x ⊑ y ∧ y ⊑ z0 → x ⊑ z0 ⊦ a0 ' + y0 ⊑ a ' + z0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list