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Proof step 6 from proof div-twoodd-05 (previous step)

0 < j, 0 ≤ j ⊦ 1 / (2 * j) = 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] → [2 * j, 1] flags: rewrite, current

Additional information

Used simplifier rules: nummult-ls-02: ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);%abs-eqall: ⊦ abs(i * j) = abs(i) * abs(j); from specification int-mult%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: 2 * j, 1