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_eval1(TRUE, x, y) → eval(-@z(x, 1@z), y)
eval(x, y) → Cond_eval5(&&(&&(>@z(x, 0@z), >=@z(0@z, x)), >=@z(0@z, y)), x, y)
Cond_eval5(TRUE, x, y) → eval(x, y)
eval(x, y) → Cond_eval4(&&(>@z(y, 0@z), >@z(x, 0@z)), x, y)
Cond_eval4(TRUE, x, y) → eval(-@z(x, 1@z), y)
eval(x, y) → Cond_eval3(&&(>@z(y, 0@z), >=@z(0@z, x)), x, y)
Cond_eval(TRUE, x, y) → eval(x, y)
Cond_eval2(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval2(&&(&&(>@z(x, 0@z), >=@z(0@z, x)), >@z(y, 0@z)), x, y)
eval(x, y) → Cond_eval1(>@z(x, 0@z), x, y)
Cond_eval3(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval(&&(&&(>@z(y, 0@z), >=@z(0@z, x)), >=@z(0@z, y)), x, y)

The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_eval1(TRUE, x, y) → eval(-@z(x, 1@z), y)
eval(x, y) → Cond_eval5(&&(&&(>@z(x, 0@z), >=@z(0@z, x)), >=@z(0@z, y)), x, y)
Cond_eval5(TRUE, x, y) → eval(x, y)
eval(x, y) → Cond_eval4(&&(>@z(y, 0@z), >@z(x, 0@z)), x, y)
Cond_eval4(TRUE, x, y) → eval(-@z(x, 1@z), y)
eval(x, y) → Cond_eval3(&&(>@z(y, 0@z), >=@z(0@z, x)), x, y)
Cond_eval(TRUE, x, y) → eval(x, y)
Cond_eval2(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval2(&&(&&(>@z(x, 0@z), >=@z(0@z, x)), >@z(y, 0@z)), x, y)
eval(x, y) → Cond_eval1(>@z(x, 0@z), x, y)
Cond_eval3(TRUE, x, y) → eval(x, -@z(y, 1@z))
eval(x, y) → Cond_eval(&&(&&(>@z(y, 0@z), >=@z(0@z, x)), >=@z(0@z, y)), x, y)

The integer pair graph contains the following rules and edges:

(0): COND_EVAL(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(x[2], 0@z), >=@z(0@z, x[2])), >@z(y[2], 0@z)), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(5): COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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


(0) -> (2), if ((y[0]* y[2])∧(x[0]* x[2]))


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


(0) -> (7), if ((y[0]* y[7])∧(x[0]* x[7]))


(0) -> (9), if ((y[0]* y[9])∧(x[0]* x[9]))


(0) -> (10), if ((y[0]* y[10])∧(x[0]* x[10]))


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


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(&&(&&(>@z(x[2], 0@z), >=@z(0@z, x[2])), >@z(y[2], 0@z)) →* TRUE))


(3) -> (5), if ((x[3]* x[5])∧(y[3]* y[5])∧(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])) →* TRUE))


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


(4) -> (2), if ((y[4]* y[2])∧(-@z(x[4], 1@z) →* x[2]))


(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


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


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(5) -> (1), if ((y[5]* y[1])∧(x[5]* x[1]))


(5) -> (2), if ((y[5]* y[2])∧(x[5]* x[2]))


(5) -> (3), if ((y[5]* y[3])∧(x[5]* x[3]))


(5) -> (7), if ((y[5]* y[7])∧(x[5]* x[7]))


(5) -> (9), if ((y[5]* y[9])∧(x[5]* x[9]))


(5) -> (10), if ((y[5]* y[10])∧(x[5]* x[10]))


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


(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) -> (7), if ((-@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(6) -> (9), if ((-@z(y[6], 1@z) →* y[9])∧(x[6]* x[9]))


(6) -> (10), if ((-@z(y[6], 1@z) →* y[10])∧(x[6]* x[10]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))


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


(8) -> (2), if ((y[8]* y[2])∧(-@z(x[8], 1@z) →* x[2]))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


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


(11) -> (2), if ((-@z(y[11], 1@z) →* y[2])∧(x[11]* x[2]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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): COND_EVAL(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(x[2], 0@z), >=@z(0@z, x[2])), >@z(y[2], 0@z)), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(5): COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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


(0) -> (2), if ((y[0]* y[2])∧(x[0]* x[2]))


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


(0) -> (7), if ((y[0]* y[7])∧(x[0]* x[7]))


(0) -> (9), if ((y[0]* y[9])∧(x[0]* x[9]))


(0) -> (10), if ((y[0]* y[10])∧(x[0]* x[10]))


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


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(&&(&&(>@z(x[2], 0@z), >=@z(0@z, x[2])), >@z(y[2], 0@z)) →* TRUE))


(3) -> (5), if ((x[3]* x[5])∧(y[3]* y[5])∧(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])) →* TRUE))


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


(4) -> (2), if ((y[4]* y[2])∧(-@z(x[4], 1@z) →* x[2]))


(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


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


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(5) -> (1), if ((y[5]* y[1])∧(x[5]* x[1]))


(5) -> (2), if ((y[5]* y[2])∧(x[5]* x[2]))


(5) -> (3), if ((y[5]* y[3])∧(x[5]* x[3]))


(5) -> (7), if ((y[5]* y[7])∧(x[5]* x[7]))


(5) -> (9), if ((y[5]* y[9])∧(x[5]* x[9]))


(5) -> (10), if ((y[5]* y[10])∧(x[5]* x[10]))


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


(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) -> (7), if ((-@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(6) -> (9), if ((-@z(y[6], 1@z) →* y[9])∧(x[6]* x[9]))


(6) -> (10), if ((-@z(y[6], 1@z) →* y[10])∧(x[6]* x[10]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))


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


(8) -> (2), if ((y[8]* y[2])∧(-@z(x[8], 1@z) →* x[2]))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


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


(11) -> (2), if ((-@z(y[11], 1@z) →* y[2])∧(x[11]* x[2]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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, y) → EVAL(x, y) the following chains were created:




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




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




For Pair EVAL(x, y) → COND_EVAL5(&&(&&(>@z(x, 0@z), >=@z(0@z, x)), >=@z(0@z, y)), x, y) the following chains were created:




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




For Pair COND_EVAL5(TRUE, x, y) → EVAL(x, y) the following chains were created:




For Pair COND_EVAL2(TRUE, x, y) → EVAL(x, -@z(y, 1@z)) the following chains were created:




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




For Pair COND_EVAL4(TRUE, x, y) → EVAL(-@z(x, 1@z), y) the following chains were created:




For Pair EVAL(x, y) → COND_EVAL4(&&(>@z(y, 0@z), >@z(x, 0@z)), x, y) the following chains were created:




For Pair EVAL(x, y) → COND_EVAL1(>@z(x, 0@z), 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:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL5(x1, x2, x3)) = -1 + (-1)x1   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(COND_EVAL(x1, x2, x3)) = -1 + (-1)x1   
POL(COND_EVAL4(x1, x2, x3)) = (-1)x1   
POL(FALSE) = 2   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL3(x1, x2, x3)) = (-1)x1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL2(x1, x2, x3)) = -1 + (-1)x1   
POL(EVAL(x1, x2)) = -1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

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

The following pairs are in Pbound:

COND_EVAL(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))

The following pairs are in P:

COND_EVAL(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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

I DP problem:
The following domains are used:

z

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

(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(&&(&&(>@z(x[2], 0@z), >=@z(0@z, x[2])), >@z(y[2], 0@z)), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(5): COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


(11) -> (2), if ((-@z(y[11], 1@z) →* y[2])∧(x[11]* x[2]))


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


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(5) -> (2), if ((y[5]* y[2])∧(x[5]* x[2]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(4) -> (2), if ((y[4]* y[2])∧(-@z(x[4], 1@z) →* x[2]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(5) -> (1), if ((y[5]* y[1])∧(x[5]* x[1]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


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


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(8) -> (2), if ((y[8]* y[2])∧(-@z(x[8], 1@z) →* x[2]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))


(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(5) -> (3), if ((y[5]* y[3])∧(x[5]* x[3]))


(5) -> (9), if ((y[5]* y[9])∧(x[5]* x[9]))


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


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(5) -> (7), if ((y[5]* y[7])∧(x[5]* x[7]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(3) -> (5), if ((x[3]* x[5])∧(y[3]* y[5])∧(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])) →* TRUE))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


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


(5) -> (10), if ((y[5]* y[10])∧(x[5]* x[10]))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                ↳ IDependencyGraphProof
IDP
                    ↳ IDPNonInfProof
              ↳ IDP

I DP problem:
The following domains are used:

z

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

(5): COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(5) -> (3), if ((y[5]* y[3])∧(x[5]* x[3]))


(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


(5) -> (9), if ((y[5]* y[9])∧(x[5]* x[9]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(5) -> (7), if ((y[5]* y[7])∧(x[5]* x[7]))


(3) -> (5), if ((x[3]* x[5])∧(y[3]* y[5])∧(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])) →* TRUE))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


(5) -> (10), if ((y[5]* y[10])∧(x[5]* x[10]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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(COND_EVAL5(x1, x2, x3)) = -1 + x3   
POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + x3   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL4(x1, x2, x3)) = -1 + x3   
POL(FALSE) = 1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL3(x1, x2, x3)) = -1 + x3   
POL(EVAL(x1, x2)) = -1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])

The following pairs are in Pbound:

COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])

The following pairs are in P:

EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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

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


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

I DP problem:
The following domains are used:

z

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

(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ 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): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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)) = x3 + x2   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL4(x1, x2, x3)) = x3 + x2 + (-1)x1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = 1   
POL(COND_EVAL3(x1, x2, x3)) = -1 + x3 + x2 + (-1)x1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL(x1, x2)) = 1 + x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])

The following pairs are in Pbound:

COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])

The following pairs are in P:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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
                    ↳ IDPNonInfProof
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


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


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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(COND_EVAL3(x1, x2, x3)) = -1 + x2   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + x2   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2)) = -1 + x1   
POL(FALSE) = 0   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in P:

EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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

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


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

(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
                                ↳ AND
                                  ↳ IDP
                                    ↳ IDependencyGraphProof
                                      ↳ IDP
                                        ↳ IDPNonInfProof
                                          ↳ 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:

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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(COND_EVAL3(x1, x2, x3)) = 2 + x3 + (-1)x2 + (-1)x1   
POL(0@z) = 0   
POL(TRUE) = 2   
POL(&&(x1, x2)) = 2   
POL(EVAL(x1, x2)) = x2 + (-1)x1   
POL(FALSE) = 2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in P:

EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], 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&&(TRUE, FALSE)1
&&(FALSE, TRUE)1FALSE1


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

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ 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): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ 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:

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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(COND_EVAL3(x1, x2, x3)) = x2 + x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL4(x1, x2, x3)) = -1 + x2 + x1   
POL(EVAL(x1, x2)) = -1 + x1   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

The following pairs are in Pbound:

COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])

The following pairs are in P:

EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
                                ↳ AND
                                  ↳ IDP
                                  ↳ IDP
                                    ↳ IDependencyGraphProof
                                      ↳ 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:

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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(COND_EVAL3(x1, x2, x3)) = -1 + (2)x3 + (-1)x2   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = 2   
POL(EVAL(x1, x2)) = (2)x2 + (-1)x1   
POL(FALSE) = 2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in P:
none

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
                                ↳ AND
                                  ↳ IDP
                                  ↳ IDP
                                    ↳ IDependencyGraphProof
                                      ↳ IDP
                                        ↳ IDPNonInfProof
                                          ↳ AND
                                            ↳ IDP
                                              ↳ IDependencyGraphProof
                                                ↳ IDP
                                                  ↳ IDPNonInfProof
                                                    ↳ AND
IDP
                                                        ↳ IDependencyGraphProof
                                                      ↳ IDP
                                            ↳ 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_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
                                ↳ AND
                                  ↳ IDP
                                  ↳ IDP
                                    ↳ IDependencyGraphProof
                                      ↳ IDP
                                        ↳ IDPNonInfProof
                                          ↳ AND
                                            ↳ IDP
                                              ↳ IDependencyGraphProof
                                                ↳ IDP
                                                  ↳ IDPNonInfProof
                                                    ↳ AND
                                                      ↳ IDP
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:

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
                                ↳ AND
                                  ↳ IDP
                                  ↳ 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:

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
                                ↳ AND
                                  ↳ IDP
                                  ↳ IDP
                                    ↳ IDependencyGraphProof
                                      ↳ 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:

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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(COND_EVAL3(x1, x2, x3)) = -1 + x3 + (-1)x2 + (-1)x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2)) = 1 + x2 + (-1)x1   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in P:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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

(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
                                ↳ AND
                                  ↳ IDP
                                  ↳ IDP
                                    ↳ IDependencyGraphProof
                                      ↳ IDP
                                        ↳ IDPNonInfProof
                                          ↳ AND
                                            ↳ IDP
                                            ↳ 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:

(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ 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:

(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


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


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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(COND_EVAL3(x1, x2, x3)) = -1 + x3   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1 + x3   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2)) = -1 + x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = 0   

The following pairs are in P>:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in P:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], 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&&(TRUE, FALSE)1
FALSE1&&(FALSE, TRUE)1


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

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])

(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ AND
                        ↳ IDP
                        ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])

(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) 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)x2   
POL(TRUE) = -1   
POL(EVAL(x1, x2)) = -1 + (2)x1   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in P:

EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])

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

-@z1


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

(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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

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(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(5): COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], -@z(y[6], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


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


(0) -> (10), if ((y[0]* y[10])∧(x[0]* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


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


(0) -> (7), if ((y[0]* y[7])∧(x[0]* x[7]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(5) -> (1), if ((y[5]* y[1])∧(x[5]* x[1]))


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


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


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


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


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


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))


(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(5) -> (3), if ((y[5]* y[3])∧(x[5]* x[3]))


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


(5) -> (9), if ((y[5]* y[9])∧(x[5]* x[9]))


(6) -> (10), if ((-@z(y[6], 1@z) →* y[10])∧(x[6]* x[10]))


(6) -> (9), if ((-@z(y[6], 1@z) →* y[9])∧(x[6]* x[9]))


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


(0) -> (9), if ((y[0]* y[9])∧(x[0]* x[9]))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(5) -> (7), if ((y[5]* y[7])∧(x[5]* x[7]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(3) -> (5), if ((x[3]* x[5])∧(y[3]* y[5])∧(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])) →* TRUE))


(6) -> (7), if ((-@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


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


(5) -> (10), if ((y[5]* y[10])∧(x[5]* x[10]))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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

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(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
(5): COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])
(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


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


(0) -> (10), if ((y[0]* y[10])∧(x[0]* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


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


(0) -> (7), if ((y[0]* y[7])∧(x[0]* x[7]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(5) -> (1), if ((y[5]* y[1])∧(x[5]* x[1]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


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


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


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))


(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


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


(5) -> (3), if ((y[5]* y[3])∧(x[5]* x[3]))


(5) -> (9), if ((y[5]* y[9])∧(x[5]* x[9]))


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


(0) -> (9), if ((y[0]* y[9])∧(x[0]* x[9]))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(5) -> (7), if ((y[5]* y[7])∧(x[5]* x[7]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(3) -> (5), if ((x[3]* x[5])∧(y[3]* y[5])∧(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])) →* TRUE))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


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


(5) -> (10), if ((y[5]* y[10])∧(x[5]* x[10]))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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[0], y[0]) → EVAL(x[0], y[0]) the following chains were created:




For Pair EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1]) the following chains were created:




For Pair COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z)) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL5(x1, x2, x3)) = -1   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3)) = -1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL4(x1, x2, x3)) = -1   
POL(COND_EVAL(x1, x2, x3)) = -1 + x1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL3(x1, x2, x3)) = x1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL(x1, x2)) = -1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])

The following pairs are in Pbound:

COND_EVAL(TRUE, x[0], y[0]) → EVAL(x[0], y[0])
COND_EVAL5(TRUE, x[5], y[5]) → EVAL(x[5], y[5])

The following pairs are in P:

EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))
EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

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
IDP
                        ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

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

(1): EVAL(x[1], y[1]) → COND_EVAL(&&(&&(>@z(y[1], 0@z), >=@z(0@z, x[1])), >=@z(0@z, y[1])), x[1], y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL5(&&(&&(>@z(x[3], 0@z), >=@z(0@z, x[3])), >=@z(0@z, y[3])), x[3], y[3])
(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (3), if ((y[4]* y[3])∧(-@z(x[4], 1@z) →* x[3]))


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


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(11) -> (3), if ((-@z(y[11], 1@z) →* y[3])∧(x[11]* x[3]))


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


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(8) -> (3), if ((y[8]* y[3])∧(-@z(x[8], 1@z) →* x[3]))


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


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(11) -> (7), if ((-@z(y[11], 1@z) →* y[7])∧(x[11]* x[7]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


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


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))


(7) -> (11), if ((x[7]* x[11])∧(y[7]* y[11])∧(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) the following chains were created:




For Pair COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 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   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL4(x1, x2, x3)) = x3 + x1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL3(x1, x2, x3)) = -1 + x3 + x1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL(x1, x2)) = -1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

The following pairs are in P:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

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
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(7): EVAL(x[7], y[7]) → COND_EVAL3(&&(>@z(y[7], 0@z), >=@z(0@z, x[7])), x[7], y[7])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(8) -> (7), if ((y[8]* y[7])∧(-@z(x[8], 1@z) →* x[7]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


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


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) 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   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(COND_EVAL4(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(EVAL(x1, x2)) = -1 + x1   
POL(FALSE) = 2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in P:

EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

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
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ 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:

(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ 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:

(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) 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)) = 1   
POL(EVAL(x1, x2)) = 1 + (2)x1   
POL(COND_EVAL4(x1, x2, x3)) = (2)x2 + x1   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])

The following pairs are in Pbound:

COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])

The following pairs are in P:

EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

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

(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])
(11): COND_EVAL3(TRUE, x[11], y[11]) → EVAL(x[11], -@z(y[11], 1@z))

(11) -> (10), if ((-@z(y[11], 1@z) →* y[10])∧(x[11]* x[10]))


(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(11) -> (9), if ((-@z(y[11], 1@z) →* y[9])∧(x[11]* x[9]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ 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:

(4): COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])
(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(4) -> (9), if ((y[4]* y[9])∧(-@z(x[4], 1@z) →* x[9]))


(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(4) -> (10), if ((y[4]* y[10])∧(-@z(x[4], 1@z) →* x[10]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(10) -> (4), if ((x[10]* x[4])∧(y[10]* y[4])∧(>@z(x[10], 0@z) →* TRUE))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4]) the following chains were created:




For Pair EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) 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   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(COND_EVAL4(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(EVAL(x1, x2)) = -1 + x1   
POL(FALSE) = 0   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[4], y[4]) → EVAL(-@z(x[4], 1@z), y[4])

The following pairs are in P:

EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

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

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


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

(10): EVAL(x[10], y[10]) → COND_EVAL1(>@z(x[10], 0@z), x[10], y[10])
(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(8) -> (10), if ((y[8]* y[10])∧(-@z(x[8], 1@z) →* x[10]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPNonInfProof
                              ↳ AND
                                ↳ IDP
                                ↳ IDP
                                  ↳ IDependencyGraphProof
                                    ↳ IDP
                                      ↳ IDPNonInfProof
                                        ↳ 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:

(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])
(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

(8) -> (9), if ((y[8]* y[9])∧(-@z(x[8], 1@z) →* x[9]))


(9) -> (8), if ((x[9]* x[8])∧(y[9]* y[8])∧(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8]) the following chains were created:




For Pair EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9]) 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(&&(x1, x2)) = -1   
POL(EVAL(x1, x2)) = 1 + x1   
POL(COND_EVAL4(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(FALSE) = 1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])

The following pairs are in Pbound:

COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])

The following pairs are in P:

COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])

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

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


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

(9): EVAL(x[9], y[9]) → COND_EVAL4(&&(>@z(y[9], 0@z), >@z(x[9], 0@z)), x[9], y[9])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, 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
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPNonInfProof
                              ↳ AND
                                ↳ IDP
                                ↳ IDP
                                  ↳ IDependencyGraphProof
                                    ↳ IDP
                                      ↳ IDPNonInfProof
                                        ↳ 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:

(8): COND_EVAL4(TRUE, x[8], y[8]) → EVAL(-@z(x[8], 1@z), y[8])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1)
eval(x0, x1)
Cond_eval5(TRUE, x0, x1)
Cond_eval4(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)


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