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Proof step 9 from proof postfix-split-stronger (previous step)

¬(x ≠ [] ∧ z ≠ []), x.last = (y + z).last, x.butlast ⊒ (y + z).butlast, ∀ x0, y, z. # x0 < # x → x0 ⊒ y + z → x0 ⊒ z ∨ (∃ x1. x1 + z = x0 ∧ x1 ⊒ y) ⊦ y + z = [], x = [], x ⊒ z, ∃ x0. x0 + z = x ∧ x0 ⊒ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
x.last = y.last, x.butlast ⊒ y.butlast, ∀ x0, y, z. # x0 < # x → x0 ⊒ y + z → x0 ⊒ z ∨ (∃ x1. x1 + z = x0 ∧ x1 ⊒ y) ⊦ x = [], y = [], x ⊒ y

Rule argument

Additional information

Used simplifier rules: postfix-empty-03: ⊦ x ⊒ [] ↔ x = [];%a: ⊦ (x + y) + z = x + y + z; from specification list%app: ⊦ x + [] = x; from specification list