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

n ≤ # x0 + 1, ∀ n. n ≤ # x0 → firstn(n, firstn(n, x0)) = firstn(n, x0) ⊦ firstn(n, firstn(n, a ' + x0)) = firstn(n, 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, firstn(n, x0)) = firstn(n, x0) ⊦ firstn(n, a ' + firstn(n -1, x0)) = a ' + firstn(n -1, 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) = []; Used terms: x0, a, n