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Proof step 4 from proof gcd-mod-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
gcd(i, j) = 1, j | i * i0 - 1, 0 < j, 1 < j, 0 < i, 1 < i, 0 ≤ i0 ⊦ ∃ i0. (i * i0) % j = 1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs