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Proof step 7 from proof gcd-conq-one-01 (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: div-neg: ⊦ i | ~ j ↔ i | j;%div-mult: ⊦ i | i * j;%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-03: ⊦ ~ i + i = 0; 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%neg-15: ⊦ i + ~ j = i - j; from specification int-abs%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs%num-subadd: ⊦ (* i) - j + (* i0) = i - i0 - j; from specification int-abs%neg-06: ⊦ 0 - i = ~ i; from specification int-abs