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Proof step 5 from proof test-div-02 (previous step)

mkDecimal(2, 0) | mkDecimal(5, 0) → div(mkDecimal(5, 0), mkDecimal(2, 0), 0) * mkDecimal(2, 0) = mkDecimal(5, 0) ⊦ div(mkDecimal(5, 0), mkDecimal(2, 0), 0) = mkDecimal(25, 1)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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