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Proof step 2 from proof first-append-01 (previous step)

n + m ≤ # x ⊦ firstn(m, restn(n, x) + y) = firstn(m, restn(n, x))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: rest-len: n ≤ # x ⊦ # restn(n, x) = # x - n;%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: y, restn(n, x), m