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Proof step 43 from proof eq-same (previous step)

¬(i→n(i / 10) < i→n(i) ∧ i→n(i0 / 10) < i→n(i0)), i % 10 = 0, mkDecimal(i / 10, abs(j) - 1).int = mkDecimal(i0 / 10, abs(j) - 1).int, mkDecimal(i / 10, abs(j) - 1).scale = mkDecimal(i0 / 10, abs(j) - 1).scale, i0 % 10 = 0, i→n(i) + i→n(i0) ≤ i→n(i / 10) + i→n(i0 / 10), ∀ i1, i2, j. i→n(i1) + i→n(i2) < i→n(i) + i→n(i0) → mkDecimal(i1, j).scale = mkDecimal(i2, j).scale ∧ mkDecimal(i1, j).int = mkDecimal(i2, j).int → i1 = i2 ⊦ j = 0, i = i0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
i % 10 = 0, mkDecimal(i / 10, abs(j) - 1).int = mkDecimal(i0 / 10, abs(j) - 1).int, mkDecimal(i / 10, abs(j) - 1).scale = mkDecimal(i0 / 10, abs(j) - 1).scale, i0 % 10 = 0, i→n(i) + i→n(i0) ≤ i→n(i / 10) + i→n(i0 / 10), ∀ i1, i2, j. i→n(i1) + i→n(i2) < i→n(i) + i→n(i0) → mkDecimal(i1, j).scale = mkDecimal(i2, j).scale ∧ mkDecimal(i1, j).int = mkDecimal(i2, j).int → i1 = i2 ⊦ j = 0, i = i0, abs(i) / 10 < abs(i) ∧ abs(i0) / 10 < abs(i0)

Rule argument

Additional information

Used simplifier rules: 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