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Proof step 6 from proof at-copy (previous step)

m0 < # ar1, ∀ m0, ar1, ar0, n, n0. m0 < # ar1 → copy(ar0, n, ar1, n0, m1)[m0] = (m0 < n0 ∨ n0 + m1 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]) ⊦ copy(ar0, n, ar1, n0, m1 +1)[m0] = (m0 < n0 ∨ n0 + m1 +1 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 < # ar1, ∀ m0, ar1, ar0, n, n0. m0 < # ar1 → copy(ar0, n, ar1, n0, m1)[m0] = (m0 < n0 ∨ n0 + m1 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]) ⊦ copy(ar0, n, ar1, n0, m1)[n0 + m1, ar0[n + m1]][m0] = (m0 < n0 ∨ n0 + m1 < m0 ⊃ ar1[m0];ar0[m0 + n - n0])

Rule argument

Additional information

Used simplifier rules: Copy-succ: ⊦ copy(ar, m, ar0, m0, n + 1) = copy(ar, m, ar0, m0, n)[m0 + n, ar[m + n]];% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)