Termination proof of /tmp/tmplKc

Termination of the given ITRSProblem could successfully be proven:

↳ ITRS

  ↳ ITRStoIDPProof

ITRS problem:
The following domains are used:

z

The TRS R consists of the following rules:

sqrtAcc(x, y) → condAcc(||(>=@z(*@z(y, y), x), <@z(y, 0@z)), x, y)
condAcc(FALSE, x, y) → sqrtAcc(x, +@z(y, 1@z))
sqrt(x) → sqrtAcc(x, 0@z)
condAcc(TRUE, x, y) → y

The set Q consists of the following terms:

sqrtAcc(x0, x1)
condAcc(FALSE, x0, x1)
sqrt(x0)
condAcc(TRUE, x0, x1)

Added dependency pairs

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

sqrtAcc(x, y) → condAcc(||(>=@z(*@z(y, y), x), <@z(y, 0@z)), x, y)
condAcc(FALSE, x, y) → sqrtAcc(x, +@z(y, 1@z))
sqrt(x) → sqrtAcc(x, 0@z)
condAcc(TRUE, x, y) → y

The integer pair graph contains the following rules and edges:

(0): SQRT(x[0]) → SQRTACC(x[0], 0@z)
(1): CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z))
(2): SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])

(0) -> (2), if ((x[0] →^* x[2]))

(1) -> (2), if ((+@z(y[1], 1@z) →^* y[2])∧(x[1] →^* x[2]))

(2) -> (1), if ((x[2] →^* x[1])∧(y[2] →^* y[1])∧(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)) →^* FALSE))

The set Q consists of the following terms:

sqrtAcc(x0, x1)
condAcc(FALSE, x0, x1)
sqrt(x0)
condAcc(TRUE, x0, x1)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(0): SQRT(x[0]) → SQRTACC(x[0], 0@z)
(1): CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z))
(2): SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])

(0) -> (2), if ((x[0] →^* x[2]))

(1) -> (2), if ((+@z(y[1], 1@z) →^* y[2])∧(x[1] →^* x[2]))

(2) -> (1), if ((x[2] →^* x[1])∧(y[2] →^* y[1])∧(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)) →^* FALSE))

The set Q consists of the following terms:

sqrtAcc(x0, x1)
condAcc(FALSE, x0, x1)
sqrt(x0)
condAcc(TRUE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(1): CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z))
(2): SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])

(2) -> (1), if ((x[2] →^* x[1])∧(y[2] →^* y[1])∧(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)) →^* FALSE))

(1) -> (2), if ((+@z(y[1], 1@z) →^* y[2])∧(x[1] →^* x[2]))

The set Q consists of the following terms:

sqrtAcc(x0, x1)
condAcc(FALSE, x0, x1)
sqrt(x0)
condAcc(TRUE, x0, x1)

The constraints were generated the following way:
The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.

For Pair CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z)) the following chains were created:

We consider the chain CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z)), SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2]), CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z)) which results in the following constraint:
(1)    (+@z(y[1], 1@z)=y[2]∧x[1]=x[2]∧x[2]=x[1]1∧||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z))=FALSE∧y[2]=y[1]1 ⇒ CONDACC(FALSE, x[1]1, y[1]1)≥NonInfC∧CONDACC(FALSE, x[1]1, y[1]1)≥SQRTACC(x[1]1, +@z(y[1]1, 1@z))∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))

We simplified constraint (1) using rules (III), (IDP_BOOLEAN) which results in the following new constraint:
(2)    (>=@z(*@z(+@z(y[1], 1@z), +@z(y[1], 1@z)), x[2])=FALSE∧<@z(+@z(y[1], 1@z), 0@z)=FALSE ⇒ CONDACC(FALSE, x[2], +@z(y[1], 1@z))≥NonInfC∧CONDACC(FALSE, x[2], +@z(y[1], 1@z))≥SQRTACC(x[2], +@z(+@z(y[1], 1@z), 1@z))∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))

We simplified constraint (2) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(3)    (-2 + x[2] + (-2)y[1] + (-1)y[1]² ≥ 0∧1 + y[1] ≥ 0 ⇒ (U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥)∧-3 + (-1)Bound + x[2] + (-2)y[1] ≥ 0∧0 ≥ 0)

We simplified constraint (3) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(4)    (-2 + x[2] + (-2)y[1] + (-1)y[1]² ≥ 0∧1 + y[1] ≥ 0 ⇒ (U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥)∧-3 + (-1)Bound + x[2] + (-2)y[1] ≥ 0∧0 ≥ 0)

We simplified constraint (4) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(5)    (-2 + x[2] + (-2)y[1] + (-1)y[1]² ≥ 0∧1 + y[1] ≥ 0 ⇒ 0 ≥ 0∧-3 + (-1)Bound + x[2] + (-2)y[1] ≥ 0∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))

We simplified constraint (5) using rule (IDP_SMT_SPLIT) which results in the following new constraint:
(6)    (x[2] ≥ 0∧1 + y[1] ≥ 0 ⇒ 0 ≥ 0∧-1 + (-1)Bound + y[1]² + x[2] ≥ 0∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))

We simplified constraint (6) using rule (IDP_SMT_SPLIT) which results in the following new constraints:
(7)    (x[2] ≥ 0∧1 + y[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧-1 + (-1)Bound + y[1]² + x[2] ≥ 0∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))

(8)    (x[2] ≥ 0∧1 + (-1)y[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧-1 + (-1)Bound + y[1]² + x[2] ≥ 0∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))

For Pair SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2]) the following chains were created:

We consider the chain SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2]) which results in the following constraint:
(9)    (SQRTACC(x[2], y[2])≥NonInfC∧SQRTACC(x[2], y[2])≥CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])∧(U^Increasing(CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])), ≥))

We simplified constraint (9) using rule (POLY_CONSTRAINTS) which results in the following new constraint:
(10)    ((U^Increasing(CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (10) using rule (IDP_POLY_SIMPLIFY) which results in the following new constraint:
(11)    ((U^Increasing(CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])), ≥)∧0 ≥ 0∧0 ≥ 0)

We simplified constraint (11) using rule (POLY_REMOVE_MIN_MAX) which results in the following new constraint:
(12)    (0 ≥ 0∧(U^Increasing(CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])), ≥)∧0 ≥ 0)

We simplified constraint (12) using rule (IDP_UNRESTRICTED_VARS) which results in the following new constraint:
(13)    (0 = 0∧0 ≥ 0∧0 = 0∧(U^Increasing(CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])), ≥)∧0 = 0∧0 = 0∧0 ≥ 0)

To summarize, we get the following constraints P_≥ for the following pairs.

CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z))
- (x[2] ≥ 0∧1 + y[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧-1 + (-1)Bound + y[1]² + x[2] ≥ 0∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))
- (x[2] ≥ 0∧1 + (-1)y[1] ≥ 0∧y[1] ≥ 0 ⇒ 0 ≥ 0∧-1 + (-1)Bound + y[1]² + x[2] ≥ 0∧(U^Increasing(SQRTACC(x[1]1, +@z(y[1]1, 1@z))), ≥))
SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])
- (0 = 0∧0 ≥ 0∧0 = 0∧(U^Increasing(CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])), ≥)∧0 = 0∧0 = 0∧0 ≥ 0)

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(>=@z(x₁, x₂)) = -1
POL(0@z) = 0
POL(SQRTACC(x₁, x₂)) = x₁ + (-2)x₂
POL(*@z(x₁, x₂)) = x₁·x₂
POL(TRUE) = 1
POL(||(x₁, x₂)) = 1
POL(+@z(x₁, x₂)) = x₁ + x₂
POL(CONDACC(x₁, x₂, x₃)) = (-1)x₁ + x₂ + (-2)x₃
POL(FALSE) = 1
POL(<@z(x₁, x₂)) = -1
POL(1@z) = 1
POL(undefined) = -1

The following pairs are in P_>:

CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z))
SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])

The following pairs are in P_bound:

CONDACC(FALSE, x[1], y[1]) → SQRTACC(x[1], +@z(y[1], 1@z))

The following pairs are in P_≥:
none

At least the following rules have been oriented under context sensitive arithmetic replacement:

||(FALSE, FALSE)¹ ↔ FALSE¹
||(TRUE, TRUE)¹ ↔ TRUE¹
||(TRUE, FALSE)¹ ↔ TRUE¹
TRUE¹ → ||(FALSE, TRUE)¹
+@z¹ ↔

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

                ↳ AND

                  ↳ IDP

                    ↳ IDependencyGraphProof

                  ↳ IDP

I DP problem:
The following domains are used:none

R is empty.
The integer pair graph is empty.
The set Q consists of the following terms:

sqrtAcc(x0, x1)
condAcc(FALSE, x0, x1)
sqrt(x0)
condAcc(TRUE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs.

↳ ITRS

  ↳ ITRStoIDPProof

    ↳ IDP

      ↳ UsableRulesProof

        ↳ IDP

          ↳ IDependencyGraphProof

            ↳ IDP

              ↳ IDPNonInfProof

                ↳ AND

                  ↳ IDP

                  ↳ IDP

                    ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(2): SQRTACC(x[2], y[2]) → CONDACC(||(>=@z(*@z(y[2], y[2]), x[2]), <@z(y[2], 0@z)), x[2], y[2])

The set Q consists of the following terms:

sqrtAcc(x0, x1)
condAcc(FALSE, x0, x1)
sqrt(x0)
condAcc(TRUE, x0, x1)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.