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Proof step 9 from proof subarray-copy-higher (previous step)

n < m1, m0 + m1 ≤ # ar, m0 + m1 ≤ n1 ⊦ copy(ar0, n0, ar, n1, n2)[n + m0] = ar[n + m0]

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:
n < m1, n + m0 < # ar, m0 + m1 ≤ n1, m0 + m1 ≤ # ar ⊦ (n + m0 < n1 ∨ n1 + n2 ≤ n + m0 ⊃ ar[n + m0];ar0[(n + m0) + n0 - n1]) = ar[n + m0]

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] → [n + m0, n2, n1, ar, n0, ar0] 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: n + m0, n2, n1, ar, n0, ar0