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Proof step 6 from proof sublist-at (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ m + n < # x → restn(n, x)[m] = x[m + n]

Rule argument

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

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: n0, x, m