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Proof step 4 from proof congmod-inverse-pos (previous step)

gcd(abs(i), abs(j)) = 1 → (∃ i0. ≡mod(abs(i) * i0, 1, abs(j))), gcd(i, j) = 1, 0 ≤ i, 0 ≤ j ⊦ ∃ i0. ≡mod(i * i0, 1, j) ∧ 0 ≤ i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
gcd(i, j) = 1, 0 ≤ i, 0 ≤ j, ≡mod(i * i0, 1, j) ⊦ ∃ i0. ≡mod(i * i0, 1, j) ∧ 0 ≤ i0

Rule argument

Additional information

Used simplifier rules: absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs