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Proof step 7 from proof postfix-split-stronger (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ x ≠ [] ∧ y ≠ [] → (x ⊒ y ↔ x.last = y.last ∧ x.butlast ⊒ y.butlast)
Proof step 9:
¬(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
Proof step 12:
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.last = z.last ∧ x.butlast ⊒ z.butlast, ∃ x0. x0 + z = x ∧ x0 ⊒ y

Rule argument

spec: list-last inst: name: postfix-rec seq: ⊦ x ≠ [] ∧ y ≠ [] → (x ⊒ y ↔ x.last = y.last ∧ x.butlast ⊒ y.butlast) subst: [y, x] → [z, x] flags: rewrite, current

Additional information

Used terms: z, x