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Proof step 15 from proof range-extend (previous step)

copy(ar, n, ar0, n0 + m0, m1)[n0 + n2] = ar1[n1 + n2], n2 < m0, range=(copy(ar, n, ar0, n0 + m0, m1), n0 + m0, ar1, n1 + m0, m1), n0 + m0 ≤ # ar0, n1 + m0 ≤ # ar1, n + m1 ≤ # ar, n0 + m0 + m1 ≤ # ar0 ⊦ ar0[n0 + n2] = ar1[n1 + n2]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + 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] → [n0 + n2, m1, n0 + m0, ar0, n, ar] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m + n < m + n0 ↔ n < n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 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) Used terms: n0 + n2, m1, n0 + m0, ar0, n, ar