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Proof step 21 from proof range-copy-drop (previous step)

n3 < m1, n0 + m0 ≤ # ar0, n + m0 ≤ # ar, n2 + m1 ≤ # ar2, n1 + m1 ≤ # ar0, ∀ n. n < m1 → ar2[n2 + n] = ar0[n1 + n], n0 + m0 ≤ n1 ∨ n1 + m1 ≤ n0 ⊦ ar2[n2 + n3] = copy(ar, n, ar0, n0, m0)[n1 + n3]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])
Proof step 23:
n3 < m1, n1 + n3 < # ar0, n1 + m1 ≤ # ar0, n2 + m1 ≤ # ar2, n + m0 ≤ # ar, n0 + m0 ≤ # ar0, ∀ n. n < m1 → ar2[n2 + n] = ar0[n1 + n], n0 + m0 ≤ n1 ∨ n1 + m1 ≤ n0 ⊦ ar2[n2 + n3] = (n1 + n3 < n0 ∨ n0 + m0 ≤ n1 + n3 ⊃ ar0[n1 + n3];ar[(n1 + n3) + n - n0])

Rule argument

spec: array-copy 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] → [n1 + n3, m0, n0, ar0, n, ar] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 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%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1 + n3, m0, n0, ar0, n, ar