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Proof step 10 from proof abs-any-def (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ i0, j. i→n(i0) < i→n(i) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ mkDecimal(abs(i), mkDecimal(i, j).scale) = mkDecimal(abs(i), j), i = 0, i % 10 = 0 ∧ j ≠ 0, j ≠ 0 ∧ i % 10 = 0

Rule argument

Additional information

Used simplifier rules: eq-zero: ⊦ mkDecimal(i, j) = mkDecimal(0, j0) ↔ i = 0; from specification decimal%scale-zero: ⊦ mkDecimal(0, j).scale = 0; from specification decimal