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

0 < i, i / gcd(i, i0) < i, i / gcd(i, i0) ≤ 0, 0 ≤ gcd(i, i0) ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: gcd(i, i0) ≠ 0

Additional information

Used simplifier rules: not-less-gcd: ⊦ 0 < i → ¬ i < gcd(i, j);%gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%gcd-zero: ⊦ gcd(0, 0) = 0;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ 0 < j → (i / j ≤ 0 ↔ i < j); (s) Used formulas: gcd(i, i0) ≠ 0