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Proof step 26 from proof test-divides-01 (previous step)

(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 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
(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 ⊦ i0 = 0

Rule argument

Additional information

Used simplifier rules: abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%abs-01: ⊦ abs(abs(i)) = abs(i); from specification int-abs