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:

eval_2(x, y) → Cond_eval_2(>=@z(0@z, x), x, y)
Cond_eval_31(TRUE, x, y) → eval_3(x, -@z(y, 1@z))
Cond_eval_11(TRUE, x, y) → eval_2(x, y)
eval_2(x, y) → Cond_eval_21(>@z(x, 0@z), x, y)
eval_3(x, y) → Cond_eval_3(>=@z(0@z, y), x, y)
eval_1(x, y) → Cond_eval_11(&&(&&(>@z(x, 0@z), >@z(y, 0@z)), >@z(x, y)), x, y)
Cond_eval_3(TRUE, x, y) → eval_1(x, y)
eval_3(x, y) → Cond_eval_31(>@z(y, 0@z), x, y)
eval_1(x, y) → Cond_eval_1(&&(&&(>@z(x, 0@z), >@z(y, 0@z)), >=@z(y, x)), x, y)
Cond_eval_21(TRUE, x, y) → eval_2(-@z(x, 1@z), y)
Cond_eval_1(TRUE, x, y) → eval_3(x, y)
Cond_eval_2(TRUE, x, y) → eval_1(x, y)

The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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:

eval_2(x, y) → Cond_eval_2(>=@z(0@z, x), x, y)
Cond_eval_31(TRUE, x, y) → eval_3(x, -@z(y, 1@z))
Cond_eval_11(TRUE, x, y) → eval_2(x, y)
eval_2(x, y) → Cond_eval_21(>@z(x, 0@z), x, y)
eval_3(x, y) → Cond_eval_3(>=@z(0@z, y), x, y)
eval_1(x, y) → Cond_eval_11(&&(&&(>@z(x, 0@z), >@z(y, 0@z)), >@z(x, y)), x, y)
Cond_eval_3(TRUE, x, y) → eval_1(x, y)
eval_3(x, y) → Cond_eval_31(>@z(y, 0@z), x, y)
eval_1(x, y) → Cond_eval_1(&&(&&(>@z(x, 0@z), >@z(y, 0@z)), >=@z(y, x)), x, y)
Cond_eval_21(TRUE, x, y) → eval_2(-@z(x, 1@z), y)
Cond_eval_1(TRUE, x, y) → eval_3(x, y)
Cond_eval_2(TRUE, x, y) → eval_1(x, y)

The integer pair graph contains the following rules and edges:

(0): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
(1): EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])
(2): EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(5): COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5])
(6): COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])
(7): EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])
(8): COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
(10): COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])
(11): EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])

(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))


(0) -> (11), if ((y[0]* y[11])∧(-@z(x[0], 1@z) →* x[11]))


(1) -> (10), if ((x[1]* x[10])∧(y[1]* y[10])∧(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])) →* TRUE))


(2) -> (5), if ((x[2]* x[5])∧(y[2]* y[5])∧(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])) →* TRUE))


(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


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


(5) -> (4), if ((y[5]* y[4])∧(x[5]* x[4]))


(5) -> (11), if ((y[5]* y[11])∧(x[5]* x[11]))


(6) -> (1), if ((y[6]* y[1])∧(x[6]* x[1]))


(6) -> (2), if ((y[6]* y[2])∧(x[6]* x[2]))


(7) -> (8), if ((x[7]* x[8])∧(y[7]* y[8])∧(>=@z(0@z, y[7]) →* TRUE))


(8) -> (1), if ((y[8]* y[1])∧(x[8]* x[1]))


(8) -> (2), if ((y[8]* y[2])∧(x[8]* x[2]))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))


(9) -> (7), if ((-@z(y[9], 1@z) →* y[7])∧(x[9]* x[7]))


(10) -> (3), if ((y[10]* y[3])∧(x[10]* x[3]))


(10) -> (7), if ((y[10]* y[7])∧(x[10]* x[7]))


(11) -> (6), if ((x[11]* x[6])∧(y[11]* y[6])∧(>=@z(0@z, x[11]) →* TRUE))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
(1): EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])
(2): EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(5): COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5])
(6): COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])
(7): EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])
(8): COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
(10): COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])
(11): EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])

(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))


(0) -> (11), if ((y[0]* y[11])∧(-@z(x[0], 1@z) →* x[11]))


(1) -> (10), if ((x[1]* x[10])∧(y[1]* y[10])∧(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])) →* TRUE))


(2) -> (5), if ((x[2]* x[5])∧(y[2]* y[5])∧(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])) →* TRUE))


(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


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


(5) -> (4), if ((y[5]* y[4])∧(x[5]* x[4]))


(5) -> (11), if ((y[5]* y[11])∧(x[5]* x[11]))


(6) -> (1), if ((y[6]* y[1])∧(x[6]* x[1]))


(6) -> (2), if ((y[6]* y[2])∧(x[6]* x[2]))


(7) -> (8), if ((x[7]* x[8])∧(y[7]* y[8])∧(>=@z(0@z, y[7]) →* TRUE))


(8) -> (1), if ((y[8]* y[1])∧(x[8]* x[1]))


(8) -> (2), if ((y[8]* y[2])∧(x[8]* x[2]))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))


(9) -> (7), if ((-@z(y[9], 1@z) →* y[7])∧(x[9]* x[7]))


(10) -> (3), if ((y[10]* y[3])∧(x[10]* x[3]))


(10) -> (7), if ((y[10]* y[7])∧(x[10]* x[7]))


(11) -> (6), if ((x[11]* x[6])∧(y[11]* y[6])∧(>=@z(0@z, x[11]) →* TRUE))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_21(TRUE, x, y) → EVAL_2(-@z(x, 1@z), y) the following chains were created:




For Pair EVAL_1(x, y) → COND_EVAL_1(&&(&&(>@z(x, 0@z), >@z(y, 0@z)), >=@z(y, x)), x, y) the following chains were created:




For Pair EVAL_1(x, y) → COND_EVAL_11(&&(&&(>@z(x, 0@z), >@z(y, 0@z)), >@z(x, y)), x, y) the following chains were created:




For Pair EVAL_3(x, y) → COND_EVAL_31(>@z(y, 0@z), x, y) the following chains were created:




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




For Pair COND_EVAL_11(TRUE, x, y) → EVAL_2(x, y) the following chains were created:




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




For Pair EVAL_3(x, y) → COND_EVAL_3(>=@z(0@z, y), x, y) the following chains were created:




For Pair COND_EVAL_3(TRUE, x, y) → EVAL_1(x, y) the following chains were created:




For Pair COND_EVAL_31(TRUE, x, y) → EVAL_3(x, -@z(y, 1@z)) the following chains were created:




For Pair COND_EVAL_1(TRUE, x, y) → EVAL_3(x, y) the following chains were created:




For Pair EVAL_2(x, y) → COND_EVAL_2(>=@z(0@z, x), x, 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(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(EVAL_1(x1, x2)) = -1 + x2 + (2)x1   
POL(FALSE) = 0   
POL(COND_EVAL_11(x1, x2, x3)) = -1 + x3 + (2)x2 + (-1)x1   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL_3(x1, x2, x3)) = -1 + x3 + (2)x2   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL_1(x1, x2, x3)) = -1 + x3 + (2)x2 + (-1)x1   
POL(EVAL_2(x1, x2)) = -1 + x2 + x1   
POL(COND_EVAL_2(x1, x2, x3)) = -1 + x3 + x2   
POL(COND_EVAL_21(x1, x2, x3)) = -1 + x3 + x2   
POL(COND_EVAL_31(x1, x2, x3)) = -1 + x3 + (2)x2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(EVAL_3(x1, x2)) = -1 + x2 + (2)x1   

The following pairs are in P>:

COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5])

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])

The following pairs are in P:

COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])
EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])
EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])
COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])
EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])

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): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
(1): EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])
(2): EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(6): COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])
(7): EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])
(8): COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
(10): COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])
(11): EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])

(10) -> (3), if ((y[10]* y[3])∧(x[10]* x[3]))


(6) -> (1), if ((y[6]* y[1])∧(x[6]* x[1]))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))


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


(6) -> (2), if ((y[6]* y[2])∧(x[6]* x[2]))


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))


(0) -> (11), if ((y[0]* y[11])∧(-@z(x[0], 1@z) →* x[11]))


(11) -> (6), if ((x[11]* x[6])∧(y[11]* y[6])∧(>=@z(0@z, x[11]) →* TRUE))


(9) -> (7), if ((-@z(y[9], 1@z) →* y[7])∧(x[9]* x[7]))


(8) -> (1), if ((y[8]* y[1])∧(x[8]* x[1]))


(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(1) -> (10), if ((x[1]* x[10])∧(y[1]* y[10])∧(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])) →* TRUE))


(10) -> (7), if ((y[10]* y[7])∧(x[10]* x[7]))


(7) -> (8), if ((x[7]* x[8])∧(y[7]* y[8])∧(>=@z(0@z, y[7]) →* TRUE))


(8) -> (2), if ((y[8]* y[2])∧(x[8]* x[2]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

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

(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(8): COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
(7): EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])
(10): COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])
(1): EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])

(1) -> (10), if ((x[1]* x[10])∧(y[1]* y[10])∧(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])) →* TRUE))


(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(10) -> (3), if ((y[10]* y[3])∧(x[10]* x[3]))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))


(10) -> (7), if ((y[10]* y[7])∧(x[10]* x[7]))


(9) -> (7), if ((-@z(y[9], 1@z) →* y[7])∧(x[9]* x[7]))


(7) -> (8), if ((x[7]* x[8])∧(y[7]* y[8])∧(>=@z(0@z, y[7]) →* TRUE))


(8) -> (1), if ((y[8]* y[1])∧(x[8]* x[1]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z)) the following chains were created:




For Pair EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3]) the following chains were created:




For Pair COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8]) the following chains were created:




For Pair EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7]) the following chains were created:




For Pair COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10]) the following chains were created:




For Pair EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1]) 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) = -1   
POL(EVAL_1(x1, x2)) = -1 + x2 + x1   
POL(&&(x1, x2)) = -1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL_3(x1, x2, x3)) = -1 + x2   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL_1(x1, x2, x3)) = -1 + x3 + x2   
POL(COND_EVAL_31(x1, x2, x3)) = x2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(EVAL_3(x1, x2)) = x1   

The following pairs are in P>:

EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])

The following pairs are in P:

COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])
EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])

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

I DP problem:
The following domains are used:

z

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

(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(8): COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
(7): EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])
(1): EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])

(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))


(9) -> (7), if ((-@z(y[9], 1@z) →* y[7])∧(x[9]* x[7]))


(7) -> (8), if ((x[7]* x[8])∧(y[7]* y[8])∧(>=@z(0@z, y[7]) →* TRUE))


(8) -> (1), if ((y[8]* y[1])∧(x[8]* x[1]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)


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
                  ↳ AND
                    ↳ IDP
                      ↳ IDPNonInfProof
                        ↳ AND
                          ↳ IDP
                            ↳ IDependencyGraphProof
IDP
                                ↳ IDPNonInfProof
                          ↳ IDP
                    ↳ 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_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3]) the following chains were created:




For Pair COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 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(TRUE) = -1   
POL(COND_EVAL_31(x1, x2, x3)) = -1 + x3   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   
POL(EVAL_3(x1, x2)) = -1 + x2   

The following pairs are in P>:

COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

The following pairs are in Pbound:

COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

The following pairs are in P:

EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])

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

-@z1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ AND
                    ↳ IDP
                      ↳ IDPNonInfProof
                        ↳ AND
                          ↳ IDP
                            ↳ IDependencyGraphProof
                              ↳ IDP
                                ↳ IDPNonInfProof
IDP
                                    ↳ IDependencyGraphProof
                          ↳ IDP
                    ↳ 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_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])


The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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
                  ↳ AND
                    ↳ 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:

(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(8): COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
(10): COND_EVAL_1(TRUE, x[10], y[10]) → EVAL_3(x[10], y[10])
(1): EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])

(1) -> (10), if ((x[1]* x[10])∧(y[1]* y[10])∧(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])) →* TRUE))


(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(10) -> (3), if ((y[10]* y[3])∧(x[10]* x[3]))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))


(8) -> (1), if ((y[8]* y[1])∧(x[8]* x[1]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)


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

(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3]) the following chains were created:




For Pair COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 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(TRUE) = 1   
POL(COND_EVAL_31(x1, x2, x3)) = 2 + (2)x3   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   
POL(EVAL_3(x1, x2)) = 2 + (2)x2   

The following pairs are in P>:

COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

The following pairs are in Pbound:

COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

The following pairs are in P:

EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])

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

-@z1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ AND
                    ↳ IDP
                      ↳ IDPNonInfProof
                        ↳ AND
                          ↳ IDP
                          ↳ IDP
                            ↳ IDependencyGraphProof
                              ↳ IDP
                                ↳ IDPNonInfProof
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_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])


The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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
                  ↳ AND
                    ↳ IDP
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:

(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(0): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])

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


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[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(0@z) = 0   
POL(EVAL_2(x1, x2)) = -1 + x1   
POL(TRUE) = 0   
POL(COND_EVAL_21(x1, x2, x3)) = -1 + x2   
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[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])

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

-@z1


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

(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])


The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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

I DP problem:
The following domains are used:

z

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

(0): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
(1): EVAL_1(x[1], y[1]) → COND_EVAL_1(&&(&&(>@z(x[1], 0@z), >@z(y[1], 0@z)), >=@z(y[1], x[1])), x[1], y[1])
(2): EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(5): COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5])
(6): COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])
(7): EVAL_3(x[7], y[7]) → COND_EVAL_3(>=@z(0@z, y[7]), x[7], y[7])
(8): COND_EVAL_3(TRUE, x[8], y[8]) → EVAL_1(x[8], y[8])
(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))
(11): EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])

(6) -> (1), if ((y[6]* y[1])∧(x[6]* x[1]))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))


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


(6) -> (2), if ((y[6]* y[2])∧(x[6]* x[2]))


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))


(0) -> (11), if ((y[0]* y[11])∧(-@z(x[0], 1@z) →* x[11]))


(11) -> (6), if ((x[11]* x[6])∧(y[11]* y[6])∧(>=@z(0@z, x[11]) →* TRUE))


(9) -> (7), if ((-@z(y[9], 1@z) →* y[7])∧(x[9]* x[7]))


(5) -> (4), if ((y[5]* y[4])∧(x[5]* x[4]))


(8) -> (1), if ((y[8]* y[1])∧(x[8]* x[1]))


(2) -> (5), if ((x[2]* x[5])∧(y[2]* y[5])∧(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])) →* TRUE))


(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(5) -> (11), if ((y[5]* y[11])∧(x[5]* x[11]))


(7) -> (8), if ((x[7]* x[8])∧(y[7]* y[8])∧(>=@z(0@z, y[7]) →* TRUE))


(8) -> (2), if ((y[8]* y[2])∧(x[8]* x[2]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ AND
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): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(11): EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])
(5): COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5])
(2): EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
(6): COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])

(2) -> (5), if ((x[2]* x[5])∧(y[2]* y[5])∧(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])) →* TRUE))


(5) -> (11), if ((y[5]* y[11])∧(x[5]* x[11]))


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


(6) -> (2), if ((y[6]* y[2])∧(x[6]* x[2]))


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))


(0) -> (11), if ((y[0]* y[11])∧(-@z(x[0], 1@z) →* x[11]))


(11) -> (6), if ((x[11]* x[6])∧(y[11]* y[6])∧(>=@z(0@z, x[11]) →* TRUE))


(5) -> (4), if ((y[5]* y[4])∧(x[5]* x[4]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0]) the following chains were created:




For Pair EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11]) the following chains were created:




For Pair COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5]) the following chains were created:




For Pair EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2]) the following chains were created:




For Pair COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6]) 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) = 0   
POL(EVAL_1(x1, x2)) = 2 + x1   
POL(&&(x1, x2)) = 0   
POL(FALSE) = -1   
POL(COND_EVAL_11(x1, x2, x3)) = 1 + x2 + (2)x1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL_2(x1, x2)) = 2   
POL(COND_EVAL_2(x1, x2, x3)) = 2   
POL(COND_EVAL_21(x1, x2, x3)) = 2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5])

The following pairs are in Pbound:

COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])

The following pairs are in P:

COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])
EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])

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
                  ↳ AND
                    ↳ 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): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(11): EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])
(5): COND_EVAL_11(TRUE, x[5], y[5]) → EVAL_2(x[5], y[5])
(2): EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])

(2) -> (5), if ((x[2]* x[5])∧(y[2]* y[5])∧(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])) →* TRUE))


(5) -> (11), if ((y[5]* y[11])∧(x[5]* x[11]))


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


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))


(0) -> (11), if ((y[0]* y[11])∧(-@z(x[0], 1@z) →* x[11]))


(5) -> (4), if ((y[5]* y[4])∧(x[5]* x[4]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)


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
                  ↳ AND
                    ↳ 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:

(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(0): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])

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


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[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(0@z) = 0   
POL(EVAL_2(x1, x2)) = (2)x1   
POL(TRUE) = -1   
POL(COND_EVAL_21(x1, x2, x3)) = -1 + (2)x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])

The following pairs are in Pbound:

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

The following pairs are in P:
none

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

-@z1


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

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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
                  ↳ AND
                    ↳ 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:

(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])


The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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
                  ↳ AND
                    ↳ 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): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(11): EVAL_2(x[11], y[11]) → COND_EVAL_2(>=@z(0@z, x[11]), x[11], y[11])
(2): EVAL_1(x[2], y[2]) → COND_EVAL_11(&&(&&(>@z(x[2], 0@z), >@z(y[2], 0@z)), >@z(x[2], y[2])), x[2], y[2])
(6): COND_EVAL_2(TRUE, x[6], y[6]) → EVAL_1(x[6], y[6])

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


(6) -> (2), if ((y[6]* y[2])∧(x[6]* x[2]))


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))


(0) -> (11), if ((y[0]* y[11])∧(-@z(x[0], 1@z) →* x[11]))


(11) -> (6), if ((x[11]* x[6])∧(y[11]* y[6])∧(>=@z(0@z, x[11]) →* TRUE))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)


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
                  ↳ AND
                    ↳ 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:

(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
(0): COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])

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


(0) -> (4), if ((y[0]* y[4])∧(-@z(x[0], 1@z) →* x[4]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4]) the following chains were created:




For Pair COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[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(0@z) = 0   
POL(EVAL_2(x1, x2)) = 2 + (2)x1   
POL(TRUE) = 2   
POL(COND_EVAL_21(x1, x2, x3)) = 1 + (2)x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = 2   

The following pairs are in P>:

EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])
COND_EVAL_21(TRUE, x[0], y[0]) → EVAL_2(-@z(x[0], 1@z), y[0])

The following pairs are in Pbound:

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

The following pairs are in P:
none

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

-@z1


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

(4): EVAL_2(x[4], y[4]) → COND_EVAL_21(>@z(x[4], 0@z), x[4], y[4])


The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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
                  ↳ AND
                    ↳ 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:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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
                  ↳ AND
                    ↳ IDP
IDP
                      ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])
(9): COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

(3) -> (9), if ((x[3]* x[9])∧(y[3]* y[9])∧(>@z(y[3], 0@z) →* TRUE))


(9) -> (3), if ((-@z(y[9], 1@z) →* y[3])∧(x[9]* x[3]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(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_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3]) the following chains were created:




For Pair COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 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(TRUE) = -1   
POL(COND_EVAL_31(x1, x2, x3)) = -1 + x3   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   
POL(EVAL_3(x1, x2)) = -1 + x2   

The following pairs are in P>:

COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

The following pairs are in Pbound:

COND_EVAL_31(TRUE, x[9], y[9]) → EVAL_3(x[9], -@z(y[9], 1@z))

The following pairs are in P:

EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])

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

-@z1


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

(3): EVAL_3(x[3], y[3]) → COND_EVAL_31(>@z(y[3], 0@z), x[3], y[3])


The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_31(TRUE, x0, x1)
Cond_eval_11(TRUE, x0, x1)
eval_3(x0, x1)
eval_1(x0, x1)
Cond_eval_3(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)


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