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Proof step 19 from proof square-sqrt (previous step)

10 ≠ 0, ∃ j0. i = j0 * 10, 0 < i, ∀ i1, i0. i→n(i1) < i→n(i) → 0 ≤ i1 → mkDecimal(sqrt(mkDecimal(i1 * i1, 2 * abs(i0)).int), mkDecimal(i1 * i1, 2 * abs(i0)).scale / 2) = mkDecimal(i1, i0) ⊦ i = 0, i0 = 0, mkDecimal(sqrt(mkDecimal((i * i) / 10, 2 * abs(i0) - 1).int), mkDecimal((i * i) / 10, 2 * abs(i0) - 1).scale / 2) = mkDecimal(i, i0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%zero-gr: ⊦ 0 < i → (0 < i * j ↔ 0 < j); from specification int-mult%div-mult-num: j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j; from specification int-div%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%zero-eq: ⊦ i * j = 0 ↔ i = 0 ∨ j = 0; from specification int-mult