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Proof step 2 from proof sublist-first-skip (previous step)

n2 ≤ # x, n + n1 ≤ n2, n + n1 ≤ # x ⊦ firstn(n1, restn(n, firstn(n2, x))) = firstn(n1, restn(n, x))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ m ≤ n ∧ n ≤ # x → restn(m, firstn(n, x)) = firstn(n - m, restn(m, x))
Proof step 4:
n ≤ n2 ∧ n2 ≤ # x, n2 ≤ # x, n + n1 ≤ n2, n + n1 ≤ # x ⊦ firstn(n1, firstn(n2 - n, restn(n, x))) = firstn(n1, restn(n, x))

Rule argument

spec: list-del inst: name: restn-firstn-better seq: ⊦ m ≤ n ∧ n ≤ # x → restn(m, firstn(n, x)) = firstn(n - m, restn(m, x)) subst: [x, n, m] → [x, n2, n] flags: rewrite, current

Additional information

Used simplifier rules: 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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: x, n2, n