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Proof step 11 from proof divides-abs-02 (previous step)

i * i0 = abs(j) * j0 ⊦ abs(j) * j0 = j * j0, abs(j) * j0 + j * j0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-mult inst: name: eq-nonzero-01 seq: ⊦ i ≠ 0 → (j * i = j0 * i ↔ j = j0) subst: [j0, i, j] → [j, j0, abs(j)] flags: rewrite

Additional information

Used simplifier rules: abs-18: ⊦ abs(i) = i ↔ 0 ≤ i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%neg-03: ⊦ ~ i + i = 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%zero-left: ⊦ 0 * i = 0; from specification int-mult Used terms: j, j0, abs(j)