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Proof step 7 from proof mod-rec-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
i ≤ j, i < 0, j < 0 ⊦ (j - i) % j + (i - j) % j = 0

Rule argument

Additional information

Used simplifier rules: negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%neg-03: ⊦ ~ i + i = 0; from specification int-abs