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Proof step 2 from proof sum-subarray (previous step)

n ≤ # ar, m0 + m1 ≤ n ⊦ sum(subarray(ar, 0, n), m0, m1) = sum(ar, m0, m1)

Heuristic structural induction applied rule structural induction

And obtained the following new goals

Proof step 3:
m0 + m +1 ≤ n, n ≤ # ar, ∀ n, ar, m0. n ≤ # ar ∧ m0 + m ≤ n → sum(subarray(ar, 0, n), m0, m) = sum(ar, m0, m) ⊦ sum(subarray(ar, 0, n), m0, m +1) = sum(ar, m0, m +1)

Rule argument

term: m1

Additional information

Used simplifier rules: sum-zero: ⊦ sum(ar, n, 0) = 0;%nat freely generated by 0, +1;;