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Proof step 20 from proof test-div-03 (previous step)

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, mkDecimal(1, 3).scale + mkDecimal(1000, 0).scale) = mkDecimal(1, 0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ ¬(i % 10 = 0 ∧ j ≠ 0 ∧ i ≠ 0) → mkDecimal(i, j).scale = abs(j)
Proof step 22:
¬(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)

Rule argument

spec: decimal inst: name: scale-base seq: ⊦ ¬(i % 10 = 0 ∧ j ≠ 0 ∧ i ≠ 0) → mkDecimal(i, j).scale = abs(j) subst: [j, i] → [3, 1] flags: rewrite

Additional information

Used terms: 3, 1