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Proof step 3 from proof sublist-at (previous step)

n0 < n, m + n ≤ # x ⊦ firstn(n, restn(m, x))[n0] = x[m + n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ n < m ∧ m ≤ # x → firstn(m, x)[n] = x[n]
Proof step 5:
n0 < n, m + n ≤ # x, n ≤ # restn(m, x) ⊦ restn(m, x)[n0] = x[m + n0]

Rule argument

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

Additional information

Used simplifier rules: rest-len: n ≤ # x ⊦ # restn(n, x) = # x - n;% ⊦ m ≤ n ↔ ¬ n < m; (ss)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ 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%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s) Used terms: n0, restn(m, x), n