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Proof step 25 from proof test-div-03 (previous step)

¬(1000 % 10 = 0 ∧ 0 ≠ 0 ∧ 1000 ≠ 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, 3 + abs(0)) = mkDecimal(1, 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2