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Proof step 5 from proof blastN-firstN (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x)
Proof step 7:
1 + # x0 - n = 0, n ≤ 1 + # x0, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a ' + x0) = firstn(1 + # x0 - n, a ' + x0)
Proof step 10:
n ≤ 1 + # x0, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ 1 + # x0 - n = 0, butlastn(n, a ' + x0) = a ' + firstn((1 + # x0 - n)-1, x0)

Rule argument

spec: <toplevel> name: firstN-rec-01 seq: ⊦ n ≠ 0 → firstn(n, a ' + x) = a ' + firstn(n -1, x) subst: [x, a, n] → [x0, a, 1 + # x0 - n] flags: <all paths>

Additional information

Used terms: x0, a, 1 + # x0 - n