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Proof step 14 from proof firstn-butlast (previous step)

# x0 = n0 + # z, x0 = firstn(n0 - 1, x0) + a ' + z, # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(n0, x0).butlast = firstn(# x0 - 1 + # z, x0) ⊦ n0 = 0, (a0 ' + firstn(n0, x0)).butlast = a0 ' + firstn(n0 - 1, x0)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
n0 = m + 1, # x0 = n0 + # z, x0 = firstn(m, x0) + a ' + z, # z ≤ # x0, x0 = firstn(# x0 - 1 + # z, x0) + a ' + z → firstn(n0, x0).butlast = firstn(# x0 - 1 + # z, x0) ⊦ n0 = 0, (a0 ' + firstn(n0, x0)).butlast = a0 ' + firstn(m, x0)

Rule argument

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

Additional information

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