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Proof step 9 from proof lastn-app (previous step)

∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ n0 + # y = 0, lastn(n0 + # y - 1, (x + y).butlast) + (x + y).last ' = lastn(n0, x) + y

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
n0 + # y = m + 1, ∀ y0, n0, x. # y0 < # y → true → lastn(n0 + # y0, x + y0) = lastn(n0, x) + y0 ⊦ n0 + # y = 0, lastn(m, (x + y).butlast) + (x + y).last ' = lastn(n0, x) + y

Rule argument

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

Additional information

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