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Proof step 3 from proof congmod-inverse-smallest (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
gcd(i, j) = 1, ≡mod(i * i0, 1, j), 0 ≤ i0, 0 ≤ i, 0 < j ⊦ ∃ i0. ≡mod(i * i0, 1, j) ∧ 0 ≤ i0 ∧ i0 < j

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs