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Proof step 32 from proof congmod-inverse (previous step)

gcd(i, j) = 1, ~(i * i0) % abs(j) = 1, 0 < i0, i < 0, i < i0 ⊦ j | i * i0 - 1, ∃ i0. ≡mod(i * i0, 1, j)

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: j = 0

Additional information

Used simplifier rules: cong-zero: ⊦ ≡mod(i, j, 0) ↔ i = j;%gcd-zero-04: ⊦ gcd(i, 0) = abs(i); from specification int-gcd%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%div-zero-zero: ⊦ 0 | i ↔ i = 0; from specification int-gcd%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%minus-one-01: ⊦ i * ~1 = ~ i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%neg-02: ⊦ ~ ~ i = i; from specification int-abs%zero-04: ⊦ 0 < i ∧ 0 < j → 0 ≠ i + j; from specification int-basic2%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs Used formulas: j = 0