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Proof step 23 from proof abs-any-def (previous step)

abs(mkDecimal(i / 10, abs(j) - 1)) = mkDecimal(abs(i / 10), abs(j) - 1), i % 10 = 0, ∀ i0, j. i→n(i0) < i→n(i) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ 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 24:
i % 10 = 0, abs(mkDecimal(i / 10, abs(j) - 1)) = mkDecimal(abs(i) / 10, abs(j) - 1), ∀ i0, j. i→n(i0) < i→n(i) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ j = 0, mkDecimal(abs(i) / 10, abs(j) - 1) = mkDecimal(abs(i), j)

Rule argument

Additional information

Used simplifier rules: abs: j ≠ 0 ⊦ abs(i / j) = abs(i) / abs(j); from specification int-div