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Proof step 24 from proof div-mult-01 (previous step)

0 < i0, i < 0, j < 0 ⊦ (i * i0) / (j * i0) = i / j, j = 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);%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 = 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 ↔ i + j ≤ 0; (s)% ⊦ i + 0 = i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ i < j → ¬ j < i; (s) Used terms: ~ j, i0, ~ i