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Proof step 29 from proof test-divides-01 (previous step)

(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

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i1, i0] → [i / 10, abs(i0) - 1]

Additional information

Used simplifier rules: abs-abs: ⊦ j ≠ 0 → abs(abs(j) - 1) = abs(j) - 1; from specification int-abs%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs-div10-le: ⊦ abs(i) ≤ abs(i) / 10 ↔ i = 0; from specification int-div%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%eq-zero: ⊦ mkDecimal(i, j) = mkDecimal(0, j0) ↔ i = 0; from specification decimal Used terms: i / 10, abs(i0) - 1