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Proof step 30 from proof square-sqrt (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
mkDecimal(sqrt(mkDecimal(j * j, 2 * (abs(i0) - 1)).int), mkDecimal(j * j, 2 * (abs(i0) - 1)).scale / 2) = mkDecimal(j, abs(i0) - 1), 0 < j, ∀ i1, i0. i→n(i1) < i→n(j * 10) → 0 ≤ i1 → mkDecimal(sqrt(mkDecimal(i1 * i1, 2 * abs(i0)).int), mkDecimal(i1 * i1, 2 * abs(i0)).scale / 2) = mkDecimal(i1, i0) ⊦ j = 0, i0 = 0, mkDecimal(sqrt(mkDecimal(j * j, 2 * abs(i0) - 2).int), mkDecimal(j * j, 2 * abs(i0) - 2).scale / 2) = mkDecimal(j * 10, i0)

Rule argument

Additional information

Used simplifier rules: abs-abs: ⊦ j ≠ 0 → abs(abs(j) - 1) = abs(j) - 1; from specification int-abs