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Proof step 14 from proof div-twoodd-07 (previous step)

j < 0 ⊦ 1 / (j * 2) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-zero-02 seq: abs(i) < abs(j) ⊦ i / j = 0 subst: [j, i] → [j * 2, 1] flags: rewrite, current

Additional information

Used simplifier rules: abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%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%neg-07: ⊦ i < ~ j ↔ i + j < 0; from specification int-abs%modzero-02: ⊦ i < 0 ∧ j0 < j ∧ 0 < j → i * j + j0 < 0; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j * 2, 1