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Proof step 16 from proof abs-any-def (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
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)

Proof step 16 from proof abs-any-def (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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