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Proof step 12 from proof test-divides-01 (previous step)

mkDecimal(3 * i, abs(i0)) = mkDecimal(1, 0), ∀ i1, i0. i→n(i1) < i→n(i) → mkDecimal(3 * i1, abs(i0)) = mkDecimal(1, 0) → false ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ i % 10 = 0 ∧ j ≠ 0 → mkDecimal(i, j) = mkDecimal(i / 10, abs(j) - 1)
Proof step 14:
¬((3 * i) % 10 = 0 ∧ abs(i0) ≠ 0), mkDecimal(3 * i, abs(i0)) = mkDecimal(1, 0), ∀ i1, i0. i→n(i1) < i→n(i) → mkDecimal(3 * i1, abs(i0)) = mkDecimal(1, 0) → false ⊦
Proof step 26:
(3 * i) % 10 = 0 ∧ abs(i0) ≠ 0, mkDecimal((3 * i) / 10, abs(abs(i0)) - 1) = mkDecimal(1, 0), ∀ i1, i0. i→n(i1) < i→n(i) → mkDecimal(3 * i1, abs(i0)) = mkDecimal(1, 0) → false ⊦

Rule argument

spec: decimal inst: name: mkDecimal-rec seq: ⊦ i % 10 = 0 ∧ j ≠ 0 → mkDecimal(i, j) = mkDecimal(i / 10, abs(j) - 1) subst: [j, i] → [abs(i0), 3 * i] flags: rewrite

Additional information

Used terms: abs(i0), 3 * i