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Proof step 11 from proof subarray-longer-eq-eq (previous step)

n0 < n, n0 < # mkarray(n), ∀ n1. n1 < n + m → copy(ar0, 0, mkarray(n + m), 0, n + m)[n1] = copy(ar1, 0, mkarray(n + m), 0, n + m)[n1] ⊦ (n0 < 0 ∨ 0 + n ≤ n0 ⊃ mkarray(n)[n0];ar0[n0 + 0 - 0]) = copy(ar1, 0, mkarray(n), 0, n)[n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n0 < n, ∀ n1. n1 < n + m → copy(ar0, 0, mkarray(n + m), 0, n + m)[n1] = copy(ar1, 0, mkarray(n + m), 0, n + m)[n1] ⊦ ar0[n0] = ar1[n0]

Rule argument

Additional information

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