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Proof step 28 from proof equality-any (previous step)

∀ 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 % 10 = 0, (i0 * 10 ^ abs(j)) % 10 = 0, j = 0, i0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0
Proof step 30:
¬(10 ^ abs(j) % 10 = 0 ∧ 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 % 10 = 0, (i0 * 10 ^ abs(j)) % 10 = 0, j = 0, i0 = 0

Rule argument

spec: int-div inst: name: mod-zero-num seq: i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0 subst: [j, i0, i] → [10, 10 ^ abs(j), i0] flags: rewrite

Additional information

Used terms: 10, 10 ^ abs(j), i0