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Proof step 4 from proof subarray-up-same (previous step)

n < # ar ⊦ # copy(ar[n, a], 0, mkarray(n), 0, n) = # copy(ar, 0, mkarray(n), 0, n) ∧ (∀ n0. n0 < # copy(ar[n, a], 0, mkarray(n), 0, n) → copy(ar[n, a], 0, mkarray(n), 0, n)[n0] = copy(ar, 0, mkarray(n), 0, n)[n0]), # ar = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < # ar, n0 < n, n < # ar, n0 + 1 < # ar ⊦ n = 0, copy(ar[n, a], 0, mkarray(n), 0, n)[n0] = copy(ar, 0, mkarray(n), 0, n)[n0], # ar = 0

Rule argument

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%Size-mkarray: ⊦ # mkarray(n) = n; from specification array