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Proof step 10 from proof square-sqrt (previous step)

¬ ¬((i * i) % 10 = 0 ∧ 2 * abs(i0) ≠ 0 ∧ i * i ≠ 0), 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) ⊦ mkDecimal(sqrt(mkDecimal(i * i, 2 * abs(i0)).int), mkDecimal(i * i, 2 * abs(i0)).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%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs%two-zero: ⊦ 2 * i = 0 ↔ i = 0; from specification int-mult%square-zero: ⊦ i * i = 0 ↔ i = 0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%mod-square-10: ⊦ (i * i) % 10 = 0 ↔ i % 10 = 0; from specification int-gcd