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Proof step 3 from proof copy-other-weak (previous step)

m0 + n0 ≤ m, m + n ≤ # ar, n0 ≤ # ar1, n ≤ # ar0 ⊦ # copy(copy(ar0, 0, ar, m, n), m0, ar1, 0, n0) = # copy(ar, m0, ar1, 0, n0) ∧ (∀ n1. n1 < # copy(copy(ar0, 0, ar, m, n), m0, ar1, 0, n0) → copy(copy(ar0, 0, ar, m, n), m0, ar1, 0, n0)[n1] = copy(ar, m0, ar1, 0, n0)[n1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n1 < # ar1, n ≤ # ar0, n0 ≤ # ar1, m + n ≤ # ar, m0 + n0 ≤ m ⊦ (n0 ≤ n1 ⊃ ar1[n1];copy(ar0, 0, ar, m, n)[n1 + m0]) = (n0 ≤ n1 ⊃ ar1[n1];ar[n1 + m0])

Rule argument

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]);%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%sub-base: ⊦ m - 0 = m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%zero-01: ⊦ ¬ m < 0; from specification nat-basic1