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Proof step 41 from proof Extsimple (previous step)

j1 / gcd(i1, j1) = j2 / gcd(i2, j2), 0 < i1, j1 < 0, j2 < 0, i2 < 0, i1 / gcd(i1, j1) < 0 ⊦ i1 = 0, i2 = 0, j1 = 0, j2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 42:
⊦ 0 < j ∧ 0 ≤ i → (i / j < j0 ↔ i < j * j0)

Rule argument

seq: ⊦ 0 < j ∧ 0 ≤ i → (i / (* j) < j0 ↔ i < j * j0) subst: [j0, j, i] → [0, gcd(i1, j1), i1] rotate: F spec: int-div inst: name: num-ls

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0;% ⊦ gcd(i, j) ≤ 0 ↔ gcd(i, j) = 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i * 0 = 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i < j → ¬ j < i; Used terms: 0, gcd(i1, j1), i1