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Proof step 2 from proof firstn-last (previous step)

m ≤ # x ⊦ m = 0, firstn(m, x).last = x[m - 1]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
m = m0 + 1, m ≤ # x ⊦ m = 0, firstn(m, x).last = x[m0]

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0