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Proof step 13 from proof gcd-one-eq (previous step)

j2 / gcd(i2, j2) = 1, j2 < 0, 0 ≤ gcd(i2, j2) ⊦ gcd(i2, j2) = j2

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 0 < gcd(i2, j2)

Additional information

Used simplifier rules: gcd-less-eq-zero: ⊦ gcd(i, j) ≤ 0 ↔ gcd(i, j) = 0;%gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%gcd-zero: ⊦ gcd(0, 0) = 0;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used formulas: 0 < gcd(i2, j2)