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Proof step 18 from proof modmod-add-law-pos (previous step)

(i + i0) % j = (i + i0 % j) % j + j, (i + i0) % j = (i0 % j + i % j) % j + 2 * j, 0 < j ⊦ j = 0

User applied rule cut formula

And obtained the following new goals

Proof step 19:
(i + i0) % j < j, (i + i0) % j = (i + i0 % j) % j + j, (i + i0) % j = (i0 % j + i % j) % j + 2 * j, 0 < j ⊦ j = 0

Rule argument

fma: (i + i0) % j < j

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j; Used formulas: (i + i0) % j < j