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Proof step 14 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

Proof step 15:
j2 / gcd(i2, j2) = 1, j2 < 0, 0 < gcd(i2, j2), 0 ≤ gcd(i2, j2) ⊦ gcd(i2, j2) = j2

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;% ⊦ i < j ↔ ¬ j ≤ i; (ss) Used formulas: 0 < gcd(i2, j2)