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Proof step 16 from proof gcd-mod-one (previous step)

i * i0 = j * i1 + 1, gcd(i, j) = 1, 1 < i, 0 < i, 1 < j, 0 < j, 0 ≤ i0, 0 ≤ i1 ⊦ (i * i0) % j = 1

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ i0 < j → (j * i + i0) % j = i0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2