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Proof step 4 from proof sublist-append-04 (previous step)

n + n0 + n1 ≤ # x0 + 1, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ firstn(n0, restn(n, a ' + x0)) + firstn(n1, restn(n + n0, a ' + x0)) = firstn(n0 + n1, restn(n, a ' + x0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < n → restn(n, x) = restn(n -1, x.rest)
Proof step 6:
¬ 0 < n, n + n0 + n1 ≤ # x0 + 1, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ firstn(n0, restn(n, a ' + x0)) + firstn(n1, restn(n + n0, a ' + x0)) = firstn(n0 + n1, restn(n, a ' + x0))
Proof step 18:
0 < n, n + n0 + n1 ≤ # x0 + 1, ∀ n, n0, n1. n + n0 + n1 ≤ # x0 → firstn(n0, restn(n, x0)) + firstn(n1, restn(n + n0, x0)) = firstn(n0 + n1, restn(n, x0)) ⊦ firstn(n0, restn(n -1, (a ' + x0).rest)) + firstn(n1, restn(n + n0, a ' + x0)) = firstn(n0 + n1, restn(n -1, (a ' + x0).rest))

Rule argument

spec: inst: name: rest-rec seq: ⊦ 0 < n → restn(n, x) = restn(n -1, x.rest) subst: [x, n] → [a ' + x0, n] flags: rewrite, current

Additional information

Used terms: a ' + x0, n