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Proof step 32 from proof test-divides-01 (previous step)

¬(i % 10 ≠ 0 ∧ gcd(3, 10) = 1 ∧ 10 ≠ 0), (3 * i) % 10 = 0, mkDecimal((3 * i) / 10, abs(i0) - 1) = mkDecimal(1, 0), ∀ i1, i0. i→n(i1) < i→n(i) → mkDecimal(3 * i1, abs(i0)) = mkDecimal(1, 0) → false ⊦ (3 * i) / 10 = 3 * i / 10, i0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 33:
(3 * i) % 10 = 0, mkDecimal((3 * i) / 10, abs(i0) - 1) = mkDecimal(1, 0), i % 10 = 0, ∀ i1, i0. i→n(i1) < i→n(i) → mkDecimal(3 * i1, abs(i0)) = mkDecimal(1, 0) → false ⊦ (3 * i) / 10 = 3 * i / 10, i0 = 0

Rule argument

Additional information

Used simplifier rules: gcd-com: ⊦ gcd(i, j) = gcd(j, i); from specification int-gcd%gcd-zero-04: ⊦ gcd(i, 0) = abs(i); from specification int-gcd%num-gcd-pos: ⊦ 0 < j ∧ j < i → gcd((* i), (* j)) = gcd(j, i % j); from specification int-gcd