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Proof step 11 from proof lastN-blastN (previous step)

n ≤ # y + 1, ∀ x0, n. # x0 < #(y + a ') → n ≤ # x0 → butlastn(n, x0) + lastn(n, x0) = x0 ⊦ butlastn(n, y + a ') + lastn(n - 1, y) = y, n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
n = m + 1, n ≤ # y + 1, ∀ x0, n. # x0 < #(y + a ') → n ≤ # x0 → butlastn(n, x0) + lastn(n, x0) = x0 ⊦ butlastn(n, y + a ') + lastn(m, y) = y, n = 0

Rule argument

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

Additional information

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