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Proof step 9 from proof kadane-correct (previous step)

max_ending_here = max_ending_here1 + a[n], max_ending_here0 = 0, max_so_far0 = 0, n = n0, n1 = 0, size = # a, n < size, n < # a, max_ending_here < 0, maximum-sum(subarray(a, 0, n), max_so_far), n ≤ size, max_ending_here1 ≤ max_so_far, 0 ≤ max_ending_here1, ∃ m. maximum-suffix-sum(subarray(a, 0, n), m, max_ending_here1) ⊦ ⟪n := n + 1⟫ ( (0 ≤ 0 ∧ 0 ≤ max_so_far ∧ n ≤ size ∧ maximum-sum(subarray(a, 0, n), max_so_far) ∧ (∃ m. maximum-suffix-sum(subarray(a, 0, n), m, 0))) ∧ size - n < size - n0)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 10:
max_ending_here = max_ending_here1 + a[n], max_ending_here0 = 0, max_so_far0 = 0, n = n0, n1 = 0, size = # a, max_ending_here < 0, n < # a, n < size, maximum-sum(subarray(a, 0, n), max_so_far), 0 ≤ max_ending_here1, max_ending_here1 ≤ max_so_far, n ≤ size, ∃ m. maximum-suffix-sum(subarray(a, 0, n), m, max_ending_here1) ⊦ (0 ≤ 0 ∧ 0 ≤ max_so_far ∧ n + 1 ≤ size ∧ maximum-sum(subarray(a, 0, n + 1), max_so_far) ∧ (∃ m. maximum-suffix-sum(subarray(a, 0, n + 1), m, 0))) ∧ size - n + 1 < size - n0

Proof step 9 from proof kadane-correct (previous step)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Rule argument

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