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Proof step 4 from proof firstn-prefix (previous step)

n ≤ # x0 + 1, ∀ n. n ≤ # x0 → firstn(n, x0) ⊑ x0 ⊦ firstn(n, a ' + x0) ⊑ a ' + x0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x)
Proof step 6:
n ≠ 0, n ≤ # x0 + 1, ∀ n. n ≤ # x0 → firstn(n, x0) ⊑ x0 ⊦ a ' + firstn(n -1, x0) ⊑ a ' + x0

Rule argument

spec: inst: name: firstN-rec-01 seq: ⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x) subst: [x, a, n] → [x0, a, n] flags: rewrite, current

Additional information

Used simplifier rules: firstN-zero: ⊦ firstn(0, x) = [];%lez: ⊦ 0 ≤ m; from specification nat%prefix-nil: ⊦ [] ⊑ x; from specification list-last Used terms: x0, a, n