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

# ar0 = n0 + n, n1 < n0 + # ar1 ⊦ (ar0 + ar1)[n + n1] = (subarray(ar0, n, n0) + ar1)[n1]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ n < # ar0 + # ar1 → (ar0 + ar1)[n] = (n < # ar0 ⊃ ar0[n];ar1[n - # ar0])
Proof step 12:
# ar0 = n0 + n, n1 < n0 + # ar1, n + n1 < # ar0 + # ar1 ⊦ (n + n1 < # ar0 ⊃ ar0[n + n1];ar1[n + n1 - # ar0]) = (subarray(ar0, n, n0) + ar1)[n1]

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 + n1, ar1, ar0] flags: <all paths>

Additional information

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