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Proof step 7 from proof lastn-mklist (previous step)

m ≤ n0 + 1, m ≤ n0, ∀ m, a. m ≤ n0 → lastn(m, mklist(a, n0)) = mklist(a, m) ⊦ lastn(m, mklist(a, n0 + 1).rest) = mklist(a, m)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [mklist(a, n0 + 1), a0, y] flags:

Additional information

Used simplifier rules: mk-empty: ⊦ mklist(a, m) = [] ↔ m = 0; from specification list-last%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: mklist(a, n0 + 1), a0, y