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Proof step 22 from proof test-div-03 (previous step)

¬(1 % 10 = 0 ∧ 3 ≠ 0 ∧ 1 ≠ 0), div(mkDecimal(1, 0), mkDecimal(1000, 0), 0) * mkDecimal(1000, 0) = mkDecimal(1, 0) ⊦ div(mkDecimal(1, 0), mkDecimal(1000, 0), 0) = mkDecimal(1, 3), mkDecimal(1000, abs(3) + mkDecimal(1000, 0).scale) = mkDecimal(1, 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
div(mkDecimal(1, 0), mkDecimal(1000, 0), 0) * mkDecimal(1000, 0) = mkDecimal(1, 0) ⊦ div(mkDecimal(1, 0), mkDecimal(1000, 0), 0) = mkDecimal(1, 3), mkDecimal(1000, 3 + mkDecimal(1000, 0).scale) = mkDecimal(1, 0)

Proof step 22 from proof test-div-03 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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