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Proof step 8 from proof mod-rec-01 (previous step)

~((j - i) % j) + (j - i) % j = 0, i < 0, j < 0, i ≤ j ⊦ ~((j - i) % j) = (i - j) % j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
j < 0, i < 0, i ≤ j ⊦ (j - i) % j + (i - j) % j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ~ i = j ↔ i + j = 0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ ~ i + i = 0; (s)