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:

Cond_eval_1(TRUE, x, y, z) → eval_2(x, y, z)
eval_2(x, y, z) → Cond_eval_21(&&(>@z(x, z), >@z(y, z)), x, y, z)
Cond_eval_21(TRUE, x, y, z) → eval_2(x, -@z(y, 1@z), z)
Cond_eval_2(TRUE, x, y, z) → eval_1(-@z(x, 1@z), y, z)
eval_1(x, y, z) → Cond_eval_1(>@z(x, z), x, y, z)
eval_2(x, y, z) → Cond_eval_2(&&(>@z(x, z), >=@z(z, y)), x, y, z)

The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


Added dependency pairs

↳ ITRS
  ↳ ITRStoIDPProof
IDP
      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

Cond_eval_1(TRUE, x, y, z) → eval_2(x, y, z)
eval_2(x, y, z) → Cond_eval_21(&&(>@z(x, z), >@z(y, z)), x, y, z)
Cond_eval_21(TRUE, x, y, z) → eval_2(x, -@z(y, 1@z), z)
Cond_eval_2(TRUE, x, y, z) → eval_1(-@z(x, 1@z), y, z)
eval_1(x, y, z) → Cond_eval_1(>@z(x, z), x, y, z)
eval_2(x, y, z) → Cond_eval_2(&&(>@z(x, z), >=@z(z, y)), x, y, z)

The integer pair graph contains the following rules and edges:

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])
(1): EVAL_1(x[1], y[1], z[1]) → COND_EVAL_1(>@z(x[1], z[1]), x[1], y[1], z[1])
(2): COND_EVAL_2(TRUE, x[2], y[2], z[2]) → EVAL_1(-@z(x[2], 1@z), y[2], z[2])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(4): EVAL_2(x[4], y[4], z[4]) → COND_EVAL_2(&&(>@z(x[4], z[4]), >=@z(z[4], y[4])), x[4], y[4], z[4])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

(0) -> (5), if ((z[0]* z[5])∧(x[0]* x[5])∧(y[0]* y[5])∧(&&(>@z(x[0], z[0]), >@z(y[0], z[0])) →* TRUE))


(1) -> (3), if ((z[1]* z[3])∧(x[1]* x[3])∧(y[1]* y[3])∧(>@z(x[1], z[1]) →* TRUE))


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


(3) -> (0), if ((y[3]* y[0])∧(z[3]* z[0])∧(x[3]* x[0]))


(3) -> (4), if ((y[3]* y[4])∧(z[3]* z[4])∧(x[3]* x[4]))


(4) -> (2), if ((z[4]* z[2])∧(x[4]* x[2])∧(y[4]* y[2])∧(&&(>@z(x[4], z[4]), >=@z(z[4], y[4])) →* TRUE))


(5) -> (0), if ((-@z(y[5], 1@z) →* y[0])∧(z[5]* z[0])∧(x[5]* x[0]))


(5) -> (4), if ((-@z(y[5], 1@z) →* y[4])∧(z[5]* z[4])∧(x[5]* x[4]))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


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
          ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])
(1): EVAL_1(x[1], y[1], z[1]) → COND_EVAL_1(>@z(x[1], z[1]), x[1], y[1], z[1])
(2): COND_EVAL_2(TRUE, x[2], y[2], z[2]) → EVAL_1(-@z(x[2], 1@z), y[2], z[2])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(4): EVAL_2(x[4], y[4], z[4]) → COND_EVAL_2(&&(>@z(x[4], z[4]), >=@z(z[4], y[4])), x[4], y[4], z[4])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

(0) -> (5), if ((z[0]* z[5])∧(x[0]* x[5])∧(y[0]* y[5])∧(&&(>@z(x[0], z[0]), >@z(y[0], z[0])) →* TRUE))


(1) -> (3), if ((z[1]* z[3])∧(x[1]* x[3])∧(y[1]* y[3])∧(>@z(x[1], z[1]) →* TRUE))


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


(3) -> (0), if ((y[3]* y[0])∧(z[3]* z[0])∧(x[3]* x[0]))


(3) -> (4), if ((y[3]* y[4])∧(z[3]* z[4])∧(x[3]* x[4]))


(4) -> (2), if ((z[4]* z[2])∧(x[4]* x[2])∧(y[4]* y[2])∧(&&(>@z(x[4], z[4]), >=@z(z[4], y[4])) →* TRUE))


(5) -> (0), if ((-@z(y[5], 1@z) →* y[0])∧(z[5]* z[0])∧(x[5]* x[0]))


(5) -> (4), if ((-@z(y[5], 1@z) →* y[4])∧(z[5]* z[4])∧(x[5]* x[4]))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


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 EVAL_2(x, y, z) → COND_EVAL_21(&&(>@z(x, z), >@z(y, z)), x, y, z) the following chains were created:




For Pair EVAL_1(x, y, z) → COND_EVAL_1(>@z(x, z), x, y, z) the following chains were created:




For Pair COND_EVAL_2(TRUE, x, y, z) → EVAL_1(-@z(x, 1@z), y, z) the following chains were created:




For Pair COND_EVAL_1(TRUE, x, y, z) → EVAL_2(x, y, z) the following chains were created:




For Pair EVAL_2(x, y, z) → COND_EVAL_2(&&(>@z(x, z), >=@z(z, y)), x, y, z) the following chains were created:




For Pair COND_EVAL_21(TRUE, x, y, z) → EVAL_2(x, -@z(y, 1@z), z) the following chains were created:




To summarize, we get the following constraints P for the following pairs.



The constraints for P> respective Pbound 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 Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL_1(x1, x2, x3, x4)) = -1 + (-1)x4 + x2   
POL(COND_EVAL_2(x1, x2, x3, x4)) = -1 + (-1)x4 + x2 + x1   
POL(TRUE) = -1   
POL(EVAL_2(x1, x2, x3)) = -1 + (-1)x3 + x1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL_21(x1, x2, x3, x4)) = -1 + (-1)x4 + x2   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL_1(x1, x2, x3)) = -1 + (-1)x3 + x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

EVAL_2(x[4], y[4], z[4]) → COND_EVAL_2(&&(>@z(x[4], z[4]), >=@z(z[4], y[4])), x[4], y[4], z[4])

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])

The following pairs are in P:

EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])
EVAL_1(x[1], y[1], z[1]) → COND_EVAL_1(>@z(x[1], z[1]), x[1], y[1], z[1])
COND_EVAL_2(TRUE, x[2], y[2], z[2]) → EVAL_1(-@z(x[2], 1@z), y[2], z[2])
COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

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

&&(FALSE, FALSE)1FALSE1
-@z1
&&(TRUE, TRUE)1TRUE1
&&(TRUE, FALSE)1FALSE1
&&(FALSE, TRUE)1FALSE1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
IDP
                ↳ IDependencyGraphProof
              ↳ IDP

I DP problem:
The following domains are used:

z

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

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])
(1): EVAL_1(x[1], y[1], z[1]) → COND_EVAL_1(>@z(x[1], z[1]), x[1], y[1], z[1])
(2): COND_EVAL_2(TRUE, x[2], y[2], z[2]) → EVAL_1(-@z(x[2], 1@z), y[2], z[2])
(4): EVAL_2(x[4], y[4], z[4]) → COND_EVAL_2(&&(>@z(x[4], z[4]), >=@z(z[4], y[4])), x[4], y[4], z[4])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

(4) -> (2), if ((z[4]* z[2])∧(x[4]* x[2])∧(y[4]* y[2])∧(&&(>@z(x[4], z[4]), >=@z(z[4], y[4])) →* TRUE))


(5) -> (0), if ((-@z(y[5], 1@z) →* y[0])∧(z[5]* z[0])∧(x[5]* x[0]))


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


(5) -> (4), if ((-@z(y[5], 1@z) →* y[4])∧(z[5]* z[4])∧(x[5]* x[4]))


(0) -> (5), if ((z[0]* z[5])∧(x[0]* x[5])∧(y[0]* y[5])∧(&&(>@z(x[0], z[0]), >@z(y[0], z[0])) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
IDP
                    ↳ IDPNonInfProof
              ↳ IDP

I DP problem:
The following domains are used:

z

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

(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])
(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])

(5) -> (0), if ((-@z(y[5], 1@z) →* y[0])∧(z[5]* z[0])∧(x[5]* x[0]))


(0) -> (5), if ((z[0]* z[5])∧(x[0]* x[5])∧(y[0]* y[5])∧(&&(>@z(x[0], z[0]), >@z(y[0], z[0])) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


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 COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5]) the following chains were created:




For Pair EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0]) the following chains were created:




To summarize, we get the following constraints P for the following pairs.



The constraints for P> respective Pbound 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 Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(TRUE) = -1   
POL(&&(x1, x2)) = 0   
POL(EVAL_2(x1, x2, x3)) = (-1)x3 + x2   
POL(COND_EVAL_21(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(FALSE) = 1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])

The following pairs are in Pbound:

COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

The following pairs are in P:

COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

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

FALSE1&&(FALSE, FALSE)1
-@z1
FALSE1&&(TRUE, FALSE)1
FALSE1&&(FALSE, TRUE)1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ AND
IDP
                          ↳ IDependencyGraphProof
                        ↳ IDP
              ↳ IDP

I DP problem:
The following domains are used:

z

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

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])


The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ AND
                        ↳ IDP
IDP
                          ↳ IDependencyGraphProof
              ↳ IDP

I DP problem:
The following domains are used:

z

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

(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])


The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])
(1): EVAL_1(x[1], y[1], z[1]) → COND_EVAL_1(>@z(x[1], z[1]), x[1], y[1], z[1])
(2): COND_EVAL_2(TRUE, x[2], y[2], z[2]) → EVAL_1(-@z(x[2], 1@z), y[2], z[2])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

(3) -> (0), if ((y[3]* y[0])∧(z[3]* z[0])∧(x[3]* x[0]))


(5) -> (0), if ((-@z(y[5], 1@z) →* y[0])∧(z[5]* z[0])∧(x[5]* x[0]))


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


(1) -> (3), if ((z[1]* z[3])∧(x[1]* x[3])∧(y[1]* y[3])∧(>@z(x[1], z[1]) →* TRUE))


(0) -> (5), if ((z[0]* z[5])∧(x[0]* x[5])∧(y[0]* y[5])∧(&&(>@z(x[0], z[0]), >@z(y[0], z[0])) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ 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:

(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])
(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])

(5) -> (0), if ((-@z(y[5], 1@z) →* y[0])∧(z[5]* z[0])∧(x[5]* x[0]))


(0) -> (5), if ((z[0]* z[5])∧(x[0]* x[5])∧(y[0]* y[5])∧(&&(>@z(x[0], z[0]), >@z(y[0], z[0])) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


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 COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5]) the following chains were created:




For Pair EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0]) the following chains were created:




To summarize, we get the following constraints P for the following pairs.



The constraints for P> respective Pbound 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 Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL_2(x1, x2, x3)) = (-1)x3 + x2   
POL(COND_EVAL_21(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + (-1)x1   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

The following pairs are in Pbound:

COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_2(x[5], -@z(y[5], 1@z), z[5])

The following pairs are in P:

EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])

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

&&(FALSE, FALSE)1FALSE1
-@z1
TRUE1&&(TRUE, TRUE)1
&&(TRUE, FALSE)1FALSE1
FALSE1&&(FALSE, TRUE)1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
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): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_21(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])


The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


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