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Proof step 11 from proof copy-rec (previous step)

n0 < # ar0 ⊦ (n0 < m0 ∨ m0 + n < n0 ⊃ ar0[n0];ar[n0 + m - m0]) = copy(ar, m + 1, ar0, m0 + 1, n)[m0, ar[m]][n0], n0 = m0 + n

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ m ≠ n → ar[n, a][m] = ar[m]

Rule argument

spec: array name: Put-other seq: ⊦ m ≠ n → ar[n, a][m] = ar[m] subst: [m, a, n, ar] → [n0, ar[m], m0, copy(ar, m + 1, ar0, m0 + 1, n)] flags: <all paths>

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%at-copy: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]);% ⊦ m ≠ n → (m < n + 1 ↔ m < n); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m + n - n + n1 = m - n1; (s)% ⊦ m = m + n ↔ n = 0; (s)% ⊦ ¬ n < n; (s)% ⊦ ¬ m + n < m; (s)% ⊦ m + n - m = n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%m < # ar ⊦ ar[m, a][m] = a; (s) Used terms: n0, ar[m], m0, copy(ar, m + 1, ar0, m0 + 1, n)