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

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: lep: ⊦ i ≤ i + j ↔ 0 ≤ j; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%mod-mult-pos-01: ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j * i + i0) % j = i0 % j; from specification int-div