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Proof step 25 from proof div-div (previous step)

(i / ~ j) / abs(j0) + ~ i / (~ j * abs(j0)) = 0, i < 0, j < 0 ⊦ (i / j) / j0 = i / (j * j0), j = 0, j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
i / (j * abs(j0)) = (i / j) / abs(j0), i < 0, j < 0 ⊦ (i / j) / j0 = i / (j * j0), j = 0, j0 = 0

Rule argument

Additional information

Used simplifier rules: Div-right: ⊦ j ≠ 0 → i / ~ j = ~(i / j);%neg-01: j ≠ 0 ⊦ ~ i / j = ~(i / j);%neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i + 0 = i; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + ~ j = i - j; (s)% ⊦ ~ ~ i = i; (s)% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; (ws)% ⊦ ~ i = 0 ↔ i = 0; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ ~ i * j = ~(i * j); (s)% ⊦ abs(i) = 0 ↔ i = 0; (s)