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Proof step 20 from proof div-nearer-zero (previous step)

~(j * i / j) = j * ~ i / j, ~ i < ~(j * i / j), i < 0, i / j * j < i, 0 < j ⊦

User applied rule insert equation

And obtained the following new goals

Proof step 21:
0 < j, i / j * j < i, i < 0, ~ i < j * ~ i / j ⊦

Rule argument

pos: left 1 flags: discard

Additional information