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Proof step 14 from proof lastn-lastn-01 (previous step)

m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0, lastn(n - 1, x.butlast)) = lastn(m0, x.butlast)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
n = m + 1, m0 < n, ∀ n, x. m0 ≤ n → lastn(m0, lastn(n, x)) = lastn(m0, x) ⊦ lastn(m0, lastn(m, x.butlast)) = lastn(m0, x.butlast), n = 0

Rule argument

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

Additional information

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