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Proof step 2 from proof div-sub (previous step)

⊦ j = 0, 1 = (j - i * j) / j + i

User applied rule cut formula

And obtained the following new goals

Proof step 3:
j - i * j = j * (1 - i) ⊦ j = 0, 1 = (j - i * j) / j + i

Rule argument

fma: j - i * j = j * (1 - i)

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%dist-rmerge: ⊦ i * j + i * j0 = i * (j + j0); from specification int-mult%Sub-def: ⊦ i + (j - i) = j; 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%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 Used formulas: j - i * j = j * (1 - i)