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Proof step 5 from proof mod-pred-03 (previous step)

j0 < j, 0 < j, 0 ≤ j0, 0 ≤ i0 ⊦ ((i0 * j + j0) + 1) % j = 0, ((i0 * j + j0) + 1) % j = j0 + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: modeq-pos seq: 0 ≤ i0 * j + j0, 0 ≤ j0, j0 < j ⊦ (i0 * j + j0) % j = j0 subst: [j0, j, i0] → [1 + j0, j, i0] flags: rewrite, current

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%zero-ge-add: ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%les: ⊦ i < j → (j ≤ i + 1 ↔ j = i + 1); from specification int-abs%succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%Mult-zero: ⊦ i * 0 = 0; 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 Used terms: 1 + j0, j, i0