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Proof step 5 from proof div-gcd-neg-not-zero (previous step)

i0 / gcd(i, i0) = 0, i0 < 0, i ≠ 0 → i0 ≠ 0 → ¬ 0 < gcd(i0, i) + i0 ⊦

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 3

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%div-gcd-zero: ⊦ i ≠ 0 → i / gcd(i, i0) ≠ 0;%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j); from specification int-div%div-same: i ≠ 0 ⊦ i / i = 1; from specification int-div