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Proof step 5 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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < # ar, n0 < n, n < # ar, n0 + 1 < # ar ⊦ copy(ar[n, a], 0, mkarray(n), 0, n)[n0] = copy(ar, 0, mkarray(n), 0, n)[n0]

Rule argument

Additional information

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