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Proof step 12 from proof subarray-append-shift-right (previous step)

n < n0 + # ar0, n < # ar0 + # ar1, n0 ≤ # ar1 ⊦ (n < # ar0 ⊃ ar0[n];ar1[n - # ar0]) = (ar0 + subarray(ar1, 0, n0))[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
n < n0 + # ar0, n < # ar0 + # ar1, n0 ≤ # ar1 ⊦ (n < # ar0 ⊃ ar0[n];ar1[n - # ar0]) = (ar0 + subarray(ar1, 0, n0))[n]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic