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Proof step 2 from proof sublist-mklist-01 (previous step)

n + m ≤ m0 ⊦ sublist(n, m, mklist(a, m0) + x) = mklist(a, m)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ n + m ≤ # x → sublist(n, m, x + y) = sublist(n, m, x)

Rule argument

spec: list-del inst: name: sublist-append seq: ⊦ n + m ≤ # x → sublist(n, m, x + y) = sublist(n, m, x) subst: [y, x, m, n] → [x, mklist(a, m0), m, n] flags: rewrite, current

Additional information

Used simplifier rules: sublist-mklist: ⊦ n + m ≤ m0 → sublist(n, m, mklist(a, m0)) = mklist(a, m);%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%mk-inj-01: ⊦ mklist(a, m) = mklist(a, n) ↔ m = n; from specification list-last%mk-len: ⊦ # mklist(a, n) = 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: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%ref-1: ⊦ ¬ n < n; from specification nat-basic1 Used terms: x, mklist(a, m0), m, n