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Proof step 11 from proof put-firstn-restn (previous step)

⊦ (a0 ' + y0 + z)[# y0, a] = a0 ' + firstn(# y0 - 1, y0 + z) + a ' + z, y0 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
# y0 = m + 1 ⊦ # y0 = 0, (a0 ' + y0 + z)[# y0, a] = a0 ' + firstn(m, y0 + z) + a ' + z, y0 = []

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [# y0, m]

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: # y0, m