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Proof step 6 from proof mult-03 (previous step)

i0 < j, 0 < i0, 0 < j, 1 < j, 0 < i, 0 < j0 ⊦ i * j - i0 = (i - 1) * j + (j - i0), (i * j - i0) / (j0 * j) = (i - 1) / j0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: dist-sub-split seq: ⊦ (i - j) * j0 = i * j0 - j * j0 subst: [j0, j, i] → [j, 1, i] flags: rewrite

Additional information

Used simplifier rules: one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%sub-add-06: ⊦ i + j - i0 + j = i - i0; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%Sub-def: ⊦ i + (j - i) = j; from specification int-abs%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult Used terms: j, 1, i