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Proof step 14 from proof congmod-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
1 < abs(j) ⊦ j = 0, ∃ i. ≡mod(~ i, 1, j)

Rule argument

Additional information

Used simplifier rules: commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%minus-one-01: ⊦ i * ~1 = ~ i; from specification int-mult%numlelsdel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) < j → (* i0) < j; from specification int-abs%gcd-com: ⊦ gcd(i, j) = gcd(j, i); from specification int-gcd%gcd-one-01: ⊦ gcd(i, 1) = 1; from specification int-gcd%abs-one-01: ⊦ i ≠ 0 ∧ i ≠ 1 → (abs(i) ≤ 1 ↔ i = ~1); from specification int-abs