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Proof step 3 from proof test-divides-03 (previous step)

⊦ ∃ dec_j0. mkDecimal(1000, 0) * dec_j0 = mkDecimal(1, 0)

User applied rule exists right

And obtained the following new goals

Proof step 4:
⊦ mkDecimal(1000, 0) * mkDecimal(1, 3) = mkDecimal(1, 0)

Rule argument

pos: right 1 terms: mkDecimal(1, 3) abort: F discard: T computed: F

Additional information

Used terms: mkDecimal(1, 3)