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Proof step 18 from proof abs-any-def (previous step)

i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ i→n(i / 10) < i→n(i), j = 0, abs(mkDecimal(i / 10, abs(j) - 1)) = mkDecimal(abs(i), j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
∀ i0, j. i→n(i0) < i→n(0) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ j = 0, abs(mkDecimal(0, abs(j) - 1)) = mkDecimal(0, j)

Rule argument

Additional information

Used simplifier rules: abs-div10-le: ⊦ abs(i) ≤ abs(i) / 10 ↔ i = 0; from specification int-div%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat