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_eval2(TRUE, x, y) → eval(-@z(x, 1@z), y)
Cond_eval(TRUE, x, y) → eval(-@z(x, 1@z), y)
Cond_eval1(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval1(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), >@z(y, x)), x, y)
eval(x, y) → Cond_eval3(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), >@z(x, y)), x, y)
Cond_eval3(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval(&&(>@z(+@z(x, y), 0@z), >@z(x, y)), x, y)
eval(x, y) → Cond_eval2(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), =@z(x, y)), x, y)

The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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:

Cond_eval2(TRUE, x, y) → eval(-@z(x, 1@z), y)
Cond_eval(TRUE, x, y) → eval(-@z(x, 1@z), y)
Cond_eval1(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval1(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), >@z(y, x)), x, y)
eval(x, y) → Cond_eval3(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), >@z(x, y)), x, y)
Cond_eval3(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval(&&(>@z(+@z(x, y), 0@z), >@z(x, y)), x, y)
eval(x, y) → Cond_eval2(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), =@z(x, y)), x, y)

The integer pair graph contains the following rules and edges:

(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(&&(>@z(+@z(x[4], y[4]), 0@z), >=@z(y[4], x[4])), >@z(x[4], y[4])), x[4], y[4])
(5): COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], -@z(y[5], 1@z))
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


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


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


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


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


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))


(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


(4) -> (5), if ((x[4]* x[5])∧(y[4]* y[5])∧(&&(&&(>@z(+@z(x[4], y[4]), 0@z), >=@z(y[4], x[4])), >@z(x[4], y[4])) →* TRUE))


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


(5) -> (2), if ((-@z(y[5], 1@z) →* y[2])∧(x[5]* x[2]))


(5) -> (3), if ((-@z(y[5], 1@z) →* y[3])∧(x[5]* x[3]))


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


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(6) -> (4), if ((-@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(7) -> (0), if ((y[7]* y[0])∧(-@z(x[7], 1@z) →* x[0]))


(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


(7) -> (3), if ((y[7]* y[3])∧(-@z(x[7], 1@z) →* x[3]))


(7) -> (4), if ((y[7]* y[4])∧(-@z(x[7], 1@z) →* x[4]))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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
          ↳ 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(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(&&(>@z(+@z(x[4], y[4]), 0@z), >=@z(y[4], x[4])), >@z(x[4], y[4])), x[4], y[4])
(5): COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], -@z(y[5], 1@z))
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


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


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


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


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


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))


(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


(4) -> (5), if ((x[4]* x[5])∧(y[4]* y[5])∧(&&(&&(>@z(+@z(x[4], y[4]), 0@z), >=@z(y[4], x[4])), >@z(x[4], y[4])) →* TRUE))


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


(5) -> (2), if ((-@z(y[5], 1@z) →* y[2])∧(x[5]* x[2]))


(5) -> (3), if ((-@z(y[5], 1@z) →* y[3])∧(x[5]* x[3]))


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


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(6) -> (4), if ((-@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(7) -> (0), if ((y[7]* y[0])∧(-@z(x[7], 1@z) →* x[0]))


(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


(7) -> (3), if ((y[7]* y[3])∧(-@z(x[7], 1@z) →* x[3]))


(7) -> (4), if ((y[7]* y[4])∧(-@z(x[7], 1@z) →* x[4]))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 EVAL(x, y) → COND_EVAL1(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), >@z(y, x)), x, y) the following chains were created:




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




For Pair EVAL(x, y) → COND_EVAL2(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), =@z(x, y)), x, y) the following chains were created:




For Pair EVAL(x, y) → COND_EVAL(&&(>@z(+@z(x, y), 0@z), >@z(x, y)), x, y) the following chains were created:




For Pair EVAL(x, y) → COND_EVAL3(&&(&&(>@z(+@z(x, y), 0@z), >=@z(y, x)), >@z(x, y)), x, y) the following chains were created:




For Pair COND_EVAL3(TRUE, x, y) → EVAL(x, -@z(y, 1@z)) the following chains were created:




For Pair COND_EVAL1(TRUE, x, y) → EVAL(x, -@z(y, 1@z)) the following chains were created:




For Pair COND_EVAL2(TRUE, x, y) → EVAL(-@z(x, 1@z), y) 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(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + x3 + (2)x2 + (-1)x1   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 0   
POL(COND_EVAL(x1, x2, x3)) = -1 + x3 + (2)x2 + (-1)x1   
POL(FALSE) = 2   
POL(>@z(x1, x2)) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL3(x1, x2, x3)) = -1 + x3 + (2)x2   
POL(COND_EVAL2(x1, x2, x3)) = -1 + x3 + (2)x2 + (-1)x1   
POL(EVAL(x1, x2)) = -1 + x2 + (2)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], -@z(y[5], 1@z))

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], -@z(y[5], 1@z))

The following pairs are in P:

EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
EVAL(x[4], y[4]) → COND_EVAL3(&&(&&(>@z(+@z(x[4], y[4]), 0@z), >=@z(y[4], x[4])), >@z(x[4], y[4])), x[4], y[4])
COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

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

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


↳ 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(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(&&(>@z(+@z(x[4], y[4]), 0@z), >=@z(y[4], x[4])), >@z(x[4], y[4])), x[4], y[4])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(7) -> (4), if ((y[7]* y[4])∧(-@z(x[7], 1@z) →* x[4]))


(6) -> (4), if ((-@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(7) -> (3), if ((y[7]* y[3])∧(-@z(x[7], 1@z) →* x[3]))


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


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


(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


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


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))


(7) -> (0), if ((y[7]* y[0])∧(-@z(x[7], 1@z) →* x[0]))


(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


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



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

↳ 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:

(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(7) -> (0), if ((y[7]* y[0])∧(-@z(x[7], 1@z) →* x[0]))


(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


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


(7) -> (3), if ((y[7]* y[3])∧(-@z(x[7], 1@z) →* x[3]))


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


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))


(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


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


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7]) the following chains were created:




For Pair EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2]) the following chains were created:




For Pair COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3]) the following chains were created:




For Pair EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@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(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(FALSE) = 0   
POL(>@z(x1, x2)) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL2(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(EVAL(x1, x2)) = x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

The following pairs are in Pbound:

COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

The following pairs are in P:

EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

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

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


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

(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


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


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


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))


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



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

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

(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


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


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


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3]) the following chains were created:




For Pair EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@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(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + x3 + x2 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3)) = -1 + x3 + x2 + (-1)x1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL(x1, x2)) = x2 + x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

The following pairs are in Pbound:

COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])

The following pairs are in P:

COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])

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

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


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

(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPNonInfProof
                              ↳ AND
                                ↳ IDP
                                  ↳ IDependencyGraphProof
IDP
                                      ↳ IDPNonInfProof
                                ↳ 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(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@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(>=@z(x1, x2)) = -1   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = (2)x3 + (2)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2)) = -1 + (2)x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

The following pairs are in P:
none

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

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


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

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPNonInfProof
                              ↳ AND
                                ↳ IDP
                                  ↳ IDependencyGraphProof
                                    ↳ IDP
                                      ↳ IDPNonInfProof
                                        ↳ AND
                                          ↳ IDP
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(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])


The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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
                      ↳ 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:

(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])

(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


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


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



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

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

(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])

(3) -> (1), if ((x[3]* x[1])∧(y[3]* y[1])∧(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])) →* TRUE))


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



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3]) 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(0@z) = 0   
POL(TRUE) = 2   
POL(&&(x1, x2)) = 0   
POL(EVAL(x1, x2)) = 1 + (2)x2 + (2)x1   
POL(COND_EVAL(x1, x2, x3)) = (2)x3 + (2)x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 0   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])
EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])

The following pairs are in Pbound:

COND_EVAL(TRUE, x[1], y[1]) → EVAL(-@z(x[1], 1@z), y[1])

The following pairs are in P:
none

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

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


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

(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])


The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPNonInfProof
                              ↳ AND
                                ↳ IDP
                                ↳ IDP
                                  ↳ IDependencyGraphProof
                                    ↳ IDP
                                      ↳ IDPNonInfProof
                                        ↳ AND
                                          ↳ IDP
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:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

↳ 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(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL(&&(>@z(+@z(x[3], y[3]), 0@z), >@z(x[3], y[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(&&(>@z(+@z(x[4], y[4]), 0@z), >=@z(y[4], x[4])), >@z(x[4], y[4])), x[4], y[4])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

(6) -> (3), if ((-@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(7) -> (4), if ((y[7]* y[4])∧(-@z(x[7], 1@z) →* x[4]))


(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(7) -> (0), if ((y[7]* y[0])∧(-@z(x[7], 1@z) →* x[0]))


(6) -> (4), if ((-@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(7) -> (3), if ((y[7]* y[3])∧(-@z(x[7], 1@z) →* x[3]))


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))


(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 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:

(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(7) -> (0), if ((y[7]* y[0])∧(-@z(x[7], 1@z) →* x[0]))


(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))


(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7]) the following chains were created:




For Pair EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2]) the following chains were created:




For Pair EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@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(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + (2)x3 + x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL2(x1, x2, x3)) = -1 + (2)x3   
POL(EVAL(x1, x2)) = -1 + (2)x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

The following pairs are in Pbound:

COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

The following pairs are in P:

COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])
EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])

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
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ 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:

(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(6) -> (2), if ((-@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

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

(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])
(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

(0) -> (6), if ((x[0]* x[6])∧(y[0]* y[6])∧(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])) →* TRUE))


(6) -> (0), if ((-@z(y[6], 1@z) →* y[0])∧(x[6]* x[0]))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@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(>=@z(x1, x2)) = -1   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + (2)x3 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2)) = 1 + (2)x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

The following pairs are in P:

COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

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

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


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

(6): COND_EVAL1(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))


The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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
                  ↳ 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:

(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])


The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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
                  ↳ 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:

(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(0): EVAL(x[0], y[0]) → COND_EVAL1(&&(&&(>@z(+@z(x[0], y[0]), 0@z), >=@z(y[0], x[0])), >@z(y[0], x[0])), x[0], y[0])

(7) -> (0), if ((y[7]* y[0])∧(-@z(x[7], 1@z) →* x[0]))


(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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

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

(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])
(7): COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

(7) -> (2), if ((y[7]* y[2])∧(-@z(x[7], 1@z) →* x[2]))


(2) -> (7), if ((x[2]* x[7])∧(y[2]* y[7])∧(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])) →* TRUE))



The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(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 EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2]) the following chains were created:




For Pair COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7]) 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(0@z) = 0   
POL(TRUE) = 2   
POL(&&(x1, x2)) = -1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(=@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL2(x1, x2, x3)) = -1 + (2)x3 + (2)x2   
POL(EVAL(x1, x2)) = -1 + (2)x2 + (2)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

The following pairs are in Pbound:

COND_EVAL2(TRUE, x[7], y[7]) → EVAL(-@z(x[7], 1@z), y[7])

The following pairs are in P:

EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])

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

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


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

(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(+@z(x[2], y[2]), 0@z), >=@z(y[2], x[2])), =@z(x[2], y[2])), x[2], y[2])


The set Q consists of the following terms:

Cond_eval2(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval3(TRUE, x0, x1)


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