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Proof step 9 from proof subarray-longer-eq-eq (previous step)

n0 < n, ∀ n1. n1 < n + m → copy(ar0, 0, mkarray(n + m), 0, n + m)[n1] = copy(ar1, 0, mkarray(n + m), 0, n + m)[n1] ⊦ copy(ar0, 0, mkarray(n), 0, n)[n0] = copy(ar1, 0, mkarray(n), 0, n)[n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])
Proof step 11:
n0 < n, n0 < # mkarray(n), ∀ n1. n1 < n + m → copy(ar0, 0, mkarray(n + m), 0, n + m)[n1] = copy(ar1, 0, mkarray(n + m), 0, n + m)[n1] ⊦ (n0 < 0 ∨ 0 + n ≤ n0 ⊃ mkarray(n)[n0];ar0[n0 + 0 - 0]) = copy(ar1, 0, mkarray(n), 0, n)[n0]

Rule argument

spec: <toplevel> name: at-copy seq: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]) subst: [m0, m, n0, ar1, n, ar0] → [n0, n, 0, mkarray(n), 0, ar0] flags: <all paths>

Additional information

Used simplifier rules: copy-exact: # ar0 = m, # ar1 = m ⊦ copy(ar0, 0, ar1, 0, m) = ar0;% ⊦ # mkarray(n) = n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: n0, n, 0, mkarray(n), 0, ar0