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Proof step 5 from proof div-mod-01 (previous step)

0 < j, 0 ≤ i0, 0 ≤ i, i0 ≤ i ⊦ (∃ i1. j * i1 = i - i0) ↔ i % j = i0 % j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, i0 ≤ i, 0 ≤ i, 0 ≤ i0 ⊦ (∃ i1. i = j * i1 + i0) ↔ i % j = i0 % j

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j; (s)