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Proof step 19 from proof congmod-inverse-pos (previous step)

¬ j ≠ 1, gcd(i, j) = 1, i0 < 0, ≡mod(i * i0, 1, j), 0 ≤ j, 0 ≤ i, j ≤ 1 ⊦ ∃ i0. ≡mod(i * i0, 1, j) ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
i0 < 0, 0 ≤ i ⊦ ∃ i1. 0 ≤ i1

Rule argument

Additional information

Used simplifier rules: cong-one: ⊦ ≡mod(i, j, 1);%gcd-one-01: ⊦ gcd(i, 1) = 1; from specification int-gcd