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Proof step 2 from proof div-gcd-is-neg (previous step)

i0 < i0 / gcd(i, i0), i0 < 0, 0 < i0 / gcd(i, i0), i0 + 1 < i0 / gcd(i, i0) ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: gcd(i, i0) ≠ 0

Additional information

Used simplifier rules: gcd-zero-zero: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used formulas: gcd(i, i0) ≠ 0