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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ' + x ⊑ y ↔ y ≠ [] ∧ a = y.first ∧ x ⊑ y.rest
Proof step 7:
y + z ≠ [] ∧ a = (y + z).first ∧ x0 ⊑ (y + z).rest, ∀ y, z. x0 ⊑ y + z → x0 ⊑ y ∨ (∃ x1. x0 = y + x1 ∧ x1 ⊑ z) ⊦ y = [], a ' + x0 ⊑ y, ∃ x. a ' + x0 = y + x ∧ x ⊑ z

Rule argument

spec: list-last inst: name: prefix-rec-01 seq: ⊦ a ' + x ⊑ y ↔ y ≠ [] ∧ a = y.first ∧ x ⊑ y.rest subst: [y, x, a] → [y + z, x0, a] flags: rewrite, current

Additional information

Used terms: y + z, x0, a