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Proof step 6 from proof mod-mult-pos-01 (previous step)

0 < j, j0 < j, 0 ≤ i, 0 ≤ j0, 0 ≤ i1 ⊦ (j * (i1 + i) + j0) % j = j0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: mod-mult-pos seq: ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ i0 < j → (j * i + i0) % j = i0 subst: [i0, i, j] → [j0, i1 + i, j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j0, i1 + i, j