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Proof step 30 from proof div-mult-01 (previous step)

0 < i0, j < 0 ⊦ (i * i0) / (j * i0) = i / j, j = 0, i < 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: j ≠ 0 ⊦ i / ~ j = ~(i / j);%Div-right: ⊦ 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 * j = 0 ↔ i = 0 ∨ j = 0; (ws)% ⊦ i * 0 = 0; (s)% ⊦ ~ i = ~ j ↔ i = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) Used terms: ~ j, i0, i