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Proof step 24 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 - 1, x0)))) = firstn(m, restn(n0 + n, a ' + x0)), n0 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
n0 = m1 + 1, 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(m1, x0)))) = firstn(m, restn(n0 + n, a ' + x0)), n0 = 0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n0, m1] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n0, m1