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Proof step 17 from proof div-mult-01 (previous step)

i0 < 0, 0 < j, 0 ≤ i ⊦ (i * i0) / (j * i0) = i / j, i0 = 0

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div-mult-02 seq: 0 < j, 0 < i0, 0 ≤ i ⊦ (i * i0) / (j * i0) = i / j subst: [j, i0, i] → [j, ~ i0, i]

Additional information

Used simplifier rules: neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);% ⊦ ~ i * j = ~(i * j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ ~ i = 0 ↔ i = 0; (s)% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; (ws)% ⊦ i * 0 = 0; (s)% ⊦ ~ ~ i = i; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ i < j → ¬ j < i; (s) Used terms: j, ~ i0, i