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Proof step 7 from proof sublist-part (previous step)

# x0 + 1 = n0 + m0, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ m0 = 0, firstn(m0 - 1, x0) + firstn(n0, restn(m0, a ' + x0)) = x0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
m0 = m + 1, # x0 + 1 = n0 + m0, ∀ n0, m0. # x0 = n0 + m0 → firstn(m0, x0) + firstn(n0, restn(m0, x0)) = x0 ⊦ m0 = 0, firstn(m, x0) + firstn(n0, restn(m0, a ' + x0)) = x0

Rule argument

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

Additional information

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