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Proof step 6 from proof scale-le (previous step)

¬(i % 10 = 0 ∧ j ≠ 0 ∧ i ≠ 0), abs(j) < mkDecimal(i, j).scale, ∀ i0, j. i→n(i0) < i→n(i) → abs(j) < mkDecimal(i0, j).scale → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
abs(j) < mkDecimal(i, j).scale, ∀ i0, j. i→n(i0) < i→n(i) → abs(j) < mkDecimal(i0, j).scale → false ⊦ i = 0, j ≠ 0 ∧ i % 10 = 0

Rule argument

Additional information

Used simplifier rules: scale-zero: ⊦ mkDecimal(0, j).scale = 0;%absls: ⊦ ¬ abs(i) < 0; from specification int-abs