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Proof step 6 from proof gcd-conq-one-01 (previous step)

1 = i * i0 + j * j0, gcd(i, j) = 1 ⊦ j | i * i0 - 1

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: i * i0 = 1 + ~ j0 * j

Additional information

Used simplifier rules: ⊦ ~ i * j = ~(i * j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i - j = i0 ↔ i = i0 + j; (s) Used formulas: i * i0 = 1 + ~ j0 * j