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Proof step 13 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), abs(j)) = mkDecimal(abs(i), j), i = 0, i % 10 = 0 ∧ j ≠ 0, j ≠ 0 ∧ i % 10 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
∀ i0, j. i→n(i0) < i→n(i) → true → abs(mkDecimal(i0, j)) = mkDecimal(abs(i0), j) ⊦ mkDecimal(abs(i), abs(j)) = mkDecimal(abs(i), j), i = 0, i % 10 = 0, j = 0

Rule argument

Additional information

Used simplifier rules: eq-same: ⊦ mkDecimal(i, j) = mkDecimal(i0, j) ↔ i = i0; from specification decimal