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Proof step 5 from proof subarray-emparray (previous step)

n2 ≤ n1 ⊦ # subarray(fill(mkarray(n0 + n1), empty), n0, n2) = # fill(mkarray(n2), empty) ∧ (∀ n. n < # subarray(fill(mkarray(n0 + n1), empty), n0, n2) → subarray(fill(mkarray(n0 + n1), empty), n0, n2)[n] = fill(mkarray(n2), empty)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < n2, n < n1, n2 ≤ n1 ⊦ fill(mkarray(n0 + n1), empty)[n0 + n] = empty

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → fill(mkarray(n), a)[m] = a; (s)% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s)% ⊦ # mkarray(n) = n; (s)% ⊦ # fill(ar, a) = # ar; (s)% ⊦ # subarray(ar, m, n) = n; (s)