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Proof step 20 from proof div-minus-onediff-01 (previous step)

1 % j = 0, 0 < j, j0 < 0 ⊦ 0 = j, j = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [j, 1] flags: rewrite, current

Additional information

Used simplifier rules: eq-one: ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = 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 Used terms: j, 1