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Proof step 22 from proof range-copy-outer (previous step)

n1 < m1, n + m0 ≤ # ar, n0 + m0 ≤ # ar0, (n3 + n0) + m1 ≤ # ar0, n3 + m1 ≤ m0, n2 + m1 ≤ # ar2, (n + n3) + m1 ≤ # ar, ∀ n4. n4 < m1 → ar2[n2 + n4] = ar[(n + n3) + n4] ⊦ ar2[n2 + n1] = copy(ar, n, ar0, n0, m0)[(n3 + n0) + n1]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])
Proof step 24:
n1 < m1, (n3 + n0) + n1 < # ar0, (n + n3) + m1 ≤ # ar, n2 + m1 ≤ # ar2, n3 + m1 ≤ m0, (n3 + n0) + m1 ≤ # ar0, n0 + m0 ≤ # ar0, n + m0 ≤ # ar, ∀ n4. n4 < m1 → ar2[n2 + n4] = ar[(n + n3) + n4] ⊦ ar2[n2 + n1] = ((n3 + n0) + n1 < n0 ∨ n0 + m0 ≤ (n3 + n0) + n1 ⊃ ar0[(n3 + n0) + n1];ar[((n3 + n0) + n1) + 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] → [(n3 + n0) + n1, 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: (n3 + n0) + n1, m0, n0, ar0, n, ar