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Proof step 10 from proof subarray-append-shift-right (previous step)

n < n0 + # ar0, n0 ≤ # ar1 ⊦ (ar0 + ar1)[n] = (ar0 + subarray(ar1, 0, n0))[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: append-at seq: ⊦ n < # ar0 + # ar1 → (ar0 + ar1)[n] = (n < # ar0 ⊃ ar0[n];ar1[n - # ar0]) subst: [n, ar1, ar0] → [n, ar1, ar0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%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 + n0 ↔ n < n0; (s)% ⊦ m < n → ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%# ar = n ⊦ subarray(ar, 0, n) = ar; (s) Used terms: n, ar1, ar0