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Proof step 14 from proof mod-add-law (previous step)

i - j * j0 = i |%| j, i0 - j * j1 = i0 |%| j, 0 < j ⊦ (i + i0) |%| j = ((i - j * j0) + (i0 - j * j1)) |%| j, j = 0

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: (i - j * j0) + (i0 - j * j1) = (i + i0) + (~ j0 - j1) * j

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%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%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%minus-left: ⊦ ~ i * j = ~(i * 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%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult Used formulas: (i - j * j0) + (i0 - j * j1) = (i + i0) + (~ j0 - j1) * j