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Proof step 2 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 inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, n, n0] flags:

Additional information

Used simplifier rules: first-mklist: ⊦ m ≤ n → firstn(m, mklist(a, n)) = mklist(a, m);%lep: ⊦ m ≤ m + n; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%mk-inj-01: ⊦ mklist(a, m) = mklist(a, n) ↔ m = n; from specification list-last%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, n, n0