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Proof step 6 from proof div-div-01 (previous step)

0 < i, i * i0 | i, 0 ≤ i0 ⊦ i = 0, i0 = 1

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] → [i, i * i0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ i ≠ 0 → (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% ⊦ 0 ≤ i → abs(i) = i; (s) Used terms: i, i * i0