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Proof step 50 from proof equality-any (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 51:
i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0
Proof step 52:
¬(10 ^ (abs(j0) - abs(j)) % 10 = 0 ∧ 10 ≠ 0), abs(j) < abs(j0), ∀ i1, i2, j2, j1. i→n(i1) + i→n(i2) < i→n(i) + i→n(i * 10 ^ (abs(j0) - abs(j))) → true → (mkDecimal(i1, j2) = mkDecimal(i2, j1) ↔ i1 * 10 ^ abs(j1) = i2 * 10 ^ abs(j2)) ⊦ i % 10 = 0, (i * 10 ^ (abs(j0) - abs(j))) % 10 = 0, i = 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(j0) - abs(j)), i] flags: rewrite

Additional information

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