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Proof step 6 from proof mod-rec-01 (previous step)

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

User applied rule insert equation

And obtained the following new goals

Proof step 7:
~((j - i) % j) + (j - i) % j = 0, i ≤ j, i < 0, j < 0 ⊦ ~((j - i) % j) = (i - j) % j

Rule argument

pos: left 1 flags: discard

Additional information