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Proof step 37 from proof copy2append0 (previous step)

# ar1 = (n2 + n1) + n0, n3 < n2 + n1, n3 < # ar0, n0 < # ar1, n3 < n1, n1 + n0 ≤ # ar1, n1 ≤ # ar0, n0 + n1 ≤ # ar1 ⊦ ar0[n3] = (subarray(ar1, 0, n0) + subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n3 + n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 38:
⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0]
Proof step 39:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 40:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 41:
# ar1 = (n2 + n1) + n0, n3 < n1, n0 < # ar1, n3 < # ar0, n3 < n2 + n1, n3 + n0 < # subarray(ar1, 0, n0) + #(subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2)), n0 + n1 ≤ # ar1, n1 ≤ # ar0, n1 + n0 ≤ # ar1, # subarray(ar1, 0, n0) ≤ n3 + n0 ⊦ ar0[n3] = (subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2))[n3 + n0 - # subarray(ar1, 0, n0)]

Rule argument

spec: <toplevel> name: append-at2 seq: ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0] subst: [n, ar1, ar0] → [n3 + n0, subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2), subarray(ar1, 0, n0)] flags: <all paths>

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ # subarray(ar, m, n) = n; (s)% ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s) Used terms: n3 + n0, subarray(ar0, 0, n1) + subarray(ar1, n0 + n1, n2), subarray(ar1, 0, n0)