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Proof step 16 from proof process-one (previous step)

((r => a0 × r0 * ls(r0, y) * del-list(rx0, n - # y + 1) * P))(H), # y < n, 1 ≤ n ⊦ r = null, r0 = null, ((del-list(r0 ' + rx0, n - 1) * P))(H -- r)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 17:
n = m + 1, ((r => a0 × r0 * ls(r0, y) * del-list(rx0, n - # y + 1) * P))(H), # y < n, 1 ≤ n ⊦ n = 0, r = null, r0 = null, ((del-list(r0 ' + rx0, m) * P))(H -- r)

Rule argument

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

Additional information

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