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Proof step 19 from proof gcd-conq-one (previous step)

1 = i * i0 + j0 * j, gcd(i, j) = 1, i0 < 0, 0 < i, j * j0 + i0 < 0, 0 ≤ j ⊦ i = 0

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: j * j0 + i0 * i ≤ j * j0 + i0

Additional information

Used simplifier rules: ⊦ i + j ≤ i0 + j ↔ i ≤ i0; (s)% ⊦ i < 0 → (i * j ≤ i ↔ 1 ≤ j); (s)% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: j * j0 + i0 * i ≤ j * j0 + i0