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Proof step 4 from proof blastN-firstN (previous step)

n ≤ # x0 + 1, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a ' + x0) = firstn(# x0 + 1 - 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:
¬ 1 + # x0 - n ≠ 0, n ≤ # x0 + 1, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a ' + x0) = firstn(# x0 + 1 - n, a ' + x0)
Proof step 9:
1 + # x0 - n ≠ 0, n ≤ # x0 + 1, ∀ n. n ≤ # x0 → butlastn(n, x0) = firstn(# x0 - n, x0) ⊦ butlastn(n, a ' + x0) = a ' + firstn((1 + # x0 - n)-1, x0)

Rule argument

spec: list-del inst: 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: rewrite, current

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: x0, a, 1 + # x0 - n