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Proof step 8 from proof copy-rec (previous step)

m0 + n < # ar0 ⊦ ar[m + n] = copy(ar, m + 1, ar0, m0 + 1, n)[m0, ar[m]][m0 + n]

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

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