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Proof step 3 from proof sublist-first-skip (previous step)

n + n1 ≤ n2, n2 ≤ # x, 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 4:
⊦ m ≤ n ∧ n ≤ # x → restn(m, firstn(n, x)) = firstn(n - m, restn(m, x))
Proof step 5:
n + n1 ≤ # x, n + n1 ≤ n2, n ≤ n2, n2 ≤ # x ⊦ firstn(n1, firstn(n2 - n, restn(n, x))) = firstn(n1, restn(n, x))

Rule argument

spec: <toplevel> 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: <all paths>

Additional information

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