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Proof step 20 from proof equality-any (previous step)

i0 % 10 = 0, ∀ i1, i2, j1, j0. i→n(i1) + i→n(i2) < i→n(i0 * 10 ^ abs(j)) + i→n(i0) → true → (mkDecimal(i1, j1) = mkDecimal(i2, j0) ↔ i1 * 10 ^ abs(j0) = i2 * 10 ^ abs(j1)) ⊦ i0 = 0, j = 0, (i0 * 10 ^ abs(j)) % 10 = 0

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, i0] flags: rewrite

Additional information

Used simplifier rules: zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%mod-zero-num: i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0; from specification int-div Used terms: 10, i0