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Proof step 10 from proof congmod-inverse (previous step)

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

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: i = 0

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i); from specification int-gcd%abs-01: ⊦ abs(abs(i)) = abs(i); from specification int-abs%gcd-com: ⊦ gcd(i, j) = gcd(j, i); from specification int-gcd%abs-06: ⊦ 0 < abs(i) ↔ i ≠ 0; from specification int-abs%Mult-zero: ⊦ i * 0 = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1 Used formulas: i = 0