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Proof step 8 from proof sublist-first-skip (previous step)

n + n1 ≤ # x, n ≤ # x, n0 + n ≤ # x, n1 ≤ n0 ⊦ firstn(n1, firstn(n0, restn(n, x))) = firstn(n1, restn(n, x))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ m ≤ n ∧ n ≤ # x → firstn(m, firstn(n, x)) = firstn(m, x)

Rule argument

spec: <toplevel> name: firstn-firstn seq: ⊦ m ≤ n ∧ n ≤ # x → firstn(m, firstn(n, x)) = firstn(m, x) subst: [x, n, m] → [restn(n, x), n0, n1] flags: <all paths>

Additional information

Used simplifier rules: rest-len: n ≤ # x ⊦ # restn(n, x) = # x - n;% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s) Used terms: restn(n, x), n0, n1