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Proof step 7 from proof gcd-mod-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
gcd(i, j) = 1, i * i0 = j * i1 + 1, 0 < j, 1 < j, 0 < i, 1 < i, 0 ≤ i0 ⊦ ∃ i0. (i * i0) % j = 1

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs