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Proof step 40 from proof square-sqrt (previous step)

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, mkDecimal(sqrt(i * i), abs(i0)) = mkDecimal(i, i0), i0 ≠ 0 ∧ i % 10 = 0, i % 10 = 0 ∧ i0 ≠ 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 41:
⊦ sqrt(i * i) = abs(i)
Proof step 42:
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, mkDecimal(abs(i), abs(i0)) = mkDecimal(i, i0), i0 ≠ 0 ∧ i % 10 = 0, i % 10 = 0 ∧ i0 ≠ 0

Rule argument

spec: int-square inst: name: inverse-ii seq: ⊦ sqrt(i * i) = abs(i) subst: [i] → [i] flags: rewrite

Additional information

Used terms: i