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Proof step 34 from proof square-sqrt (previous step)

mkDecimal(sqrt(mkDecimal(j * j, abs(i0) * 2 - 2).int), mkDecimal(j * j, abs(i0) * 2 - 2).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(j, abs(i0) - 1) = mkDecimal(j * 10, i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: decimal inst: name: mkDecimal-rec seq: ⊦ i % 10 = 0 ∧ j ≠ 0 → mkDecimal(i, j) = mkDecimal(i / 10, abs(j) - 1) subst: [j, i] → [i0, j * 10] flags: rewrite

Additional information

Used simplifier rules: mod-zero-num: i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0; from specification int-div%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%div-mult-num: j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j; from specification int-div%one: ⊦ 1 * i = i; from specification int-mult%eq-same: ⊦ mkDecimal(i, j) = mkDecimal(i0, j) ↔ i = i0; from specification decimal Used terms: i0, j * 10