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Proof step 4 from proof sublist-move (previous step)

n + m ≤ m0, n0 + m0 ≤ # x0 + 1, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(n, firstn(m0, restn(n0, a ' + x0)))) = firstn(m, restn(n0 + 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 < n0, n + m ≤ m0, n0 + m0 ≤ # x0 + 1, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(n, firstn(m0, restn(n0, a ' + x0)))) = firstn(m, restn(n0 + n, a ' + x0))
Proof step 23:
0 < n0, n + m ≤ m0, n0 + m0 ≤ # x0 + 1, ∀ n, m, m0, n0. n + m ≤ m0 ∧ n0 + m0 ≤ # x0 → firstn(m, restn(n, firstn(m0, restn(n0, x0)))) = firstn(m, restn(n0 + n, x0)) ⊦ firstn(m, restn(n, firstn(m0, restn(n0 -1, (a ' + x0).rest)))) = firstn(m, restn(n0 + n, a ' + x0))

Rule argument

spec: inst: name: rest-rec seq: ⊦ 0 < n → restn(n, x) = restn(n -1, x.rest) subst: [x, n] → [a ' + x0, n0] flags: rewrite, current

Additional information

Used terms: a ' + x0, n0