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Proof step 56 from proof equality-any (previous step)

abs(j0) < abs(j), ∀ i1, i2, j1, j2. i→n(i1) + i→n(i2) < i→n(i0 * 10 ^ (abs(j) - abs(j0))) + i→n(i0) → true → (mkDecimal(i1, j1) = mkDecimal(i2, j2) ↔ i1 * 10 ^ abs(j2) = i2 * 10 ^ abs(j1)) ⊦ (i0 * 10 ^ (abs(j) - abs(j0))) % 10 = 0, i0 % 10 = 0, i0 = 0, abs(j) = abs(j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 57:
i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0
Proof step 58:
¬(10 ^ (abs(j) - abs(j0)) % 10 = 0 ∧ 10 ≠ 0), abs(j0) < abs(j), ∀ i1, i2, j1, j2. i→n(i1) + i→n(i2) < i→n(i0 * 10 ^ (abs(j) - abs(j0))) + i→n(i0) → true → (mkDecimal(i1, j1) = mkDecimal(i2, j2) ↔ i1 * 10 ^ abs(j2) = i2 * 10 ^ abs(j1)) ⊦ (i0 * 10 ^ (abs(j) - abs(j0))) % 10 = 0, i0 % 10 = 0, i0 = 0, abs(j) = abs(j0)

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) - abs(j0)), i0] flags: rewrite

Additional information

Used terms: 10, 10 ^ (abs(j) - abs(j0)), i0