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Proof step 2 from proof div-mult-div (previous step)

i * j | j, 0 < j, 0 < i, 1 < i ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [j, i * j] flags: rewrite, current

Additional information

Used simplifier rules: eq-same-02: ⊦ 0 < i → (i * j = i ↔ j = 1); from specification int-mult%eq-one: ⊦ i * j = 1 ↔ abs(i) = 1 ∧ 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%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1 Used terms: j, i * j