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Proof step 5 from proof list-split (previous step)

# x ≤ # y0 + 1, x ⊑ (a ' + y0) + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ ∃ y. a ' + y0 = x + y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] → (x ⊑ a ' + y ↔ x.head = a ∧ x.tail ⊑ y)
Proof step 7:
x.head = a, # x ≤ # y0 + 1, x.tail ⊑ y0 + z, ∀ x, z. # x ≤ # y0 ∧ x ⊑ y0 + z → (∃ y1. y0 = x + y1) ⊦ x = [], ∃ y. a ' + y0 = x + y

Rule argument

spec: list-last name: prefix-rec seq: ⊦ x ≠ [] → (x ⊑ a ' + y ↔ x.head = a ∧ x.tail ⊑ y) subst: [y, a, x] → [y0 + z, a, x] flags: <all paths>

Additional information

Used simplifier rules: ⊦ [] ⊑ x; (s)% ⊦ # [] = 0; (s)% ⊦ 0 ≤ m; (s)% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: y0 + z, a, x