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Proof step 3 from proof sublist-mklist (previous step)

n + m ≤ m0 ⊦ firstn(m, mklist(a, m0 - n)) = mklist(a, m)

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, n, n0]

Additional information

Used simplifier rules: first-mklist: ⊦ m ≤ n → firstn(m, mklist(a, n)) = mklist(a, m);% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ mklist(a, m) = mklist(a, n) ↔ m = n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ m + n < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, n, n0