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Proof step 3 from proof div-mult-div (previous step)

0 < i, 1 < i, 0 < j, i * j | j ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [j, i * j] flags: <all paths>

Additional information

Used simplifier rules: ⊦ 0 < i → (i * j = i ↔ j = 1); (s)% ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j; (ws)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ 1 < i → abs(i) = i; (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1 Used terms: j, i * j