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Proof step 89 from proof equality-any (previous step)

i % 10 = 0, mkDecimal(i / 10, abs(j) - 1) = mkDecimal(i0, j0) ↔ i / 10 * 10 ^ abs(j0) = i0 * 10 ^ (abs(j) - 1), ∀ i1, i2, j, j0. i→n(i1) + i→n(i2) < i→n(i) + i→n(i0) → true → (mkDecimal(i1, j) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j)) ⊦ j = 0, mkDecimal(i / 10, abs(j) - 1) = mkDecimal(i0, j0) ↔ i * 10 ^ abs(j0) = i0 * 10 ^ (abs(j) - 1) * 10

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: modex seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [10, i] flags: rewrite

Additional information

Used simplifier rules: div-mult-num: j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j; from specification int-div%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%eq-nonzero: ⊦ i ≠ 0 → (i * j = i * j0 ↔ j = j0); from specification int-mult Used terms: 10, i