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Proof step 6 from proof kadane-correct (previous step)

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

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 7:
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, 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) ⊦ ⟪if max_ending_here < 0 then max_ending_here := 0 else if max_so_far < max_ending_here then max_so_far := max_ending_here; n := n + 1⟫ ( ( 0 ≤ max_ending_here ∧ max_ending_here ≤ max_so_far ∧ n ≤ size ∧ maximum-sum(subarray(a, 0, n), max_so_far) ∧ (∃ m. maximum-suffix-sum(subarray(a, 0, n), m, max_ending_here))) ∧ size - n < size - n0)

Rule argument

Additional information

Used simplifier rules: # ar0 = n ⊦ subarray(ar0, 0, n) = ar0; (s)% ⊦ m < n → m ≤ n; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)