Termination of the given ITRSProblem could successfully be proven:



ITRS
  ↳ ITRStoIDPProof

ITRS problem:
The following domains are used:

z

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


Added dependency pairs

↳ ITRS
  ↳ ITRStoIDPProof
IDP
      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

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

The integer pair graph contains the following rules and edges:

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

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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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


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


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


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


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


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


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


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


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


(5) -> (9), if ((z[5]* z[9])∧(x[5]* x[9])∧(y[5]* y[9])∧(&&(&&(>@z(y[5], z[5]), >=@z(z[5], x[5])), >=@z(z[5], y[5])) →* TRUE))


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


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


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


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


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


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


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


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


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


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


(8) -> (5), if ((-@z(y[8], 1@z) →* y[5])∧(z[8]* z[5])∧(x[8]* x[5]))


(8) -> (6), if ((-@z(y[8], 1@z) →* y[6])∧(z[8]* z[6])∧(x[8]* x[6]))


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


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


(9) -> (2), if ((y[9]* y[2])∧(z[9]* z[2])∧(x[9]* x[2]))


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


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


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


(9) -> (10), if ((y[9]* y[10])∧(z[9]* z[10])∧(x[9]* x[10]))


(10) -> (7), if ((z[10]* z[7])∧(x[10]* x[7])∧(y[10]* y[7])∧(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])) →* TRUE))


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


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


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


(11) -> (5), if ((-@z(y[11], 1@z) →* y[5])∧(z[11]* z[5])∧(x[11]* x[5]))


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
IDP
          ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

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

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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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


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


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


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


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


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


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


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


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


(5) -> (9), if ((z[5]* z[9])∧(x[5]* x[9])∧(y[5]* y[9])∧(&&(&&(>@z(y[5], z[5]), >=@z(z[5], x[5])), >=@z(z[5], y[5])) →* TRUE))


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


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


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


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


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


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


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


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


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


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


(8) -> (5), if ((-@z(y[8], 1@z) →* y[5])∧(z[8]* z[5])∧(x[8]* x[5]))


(8) -> (6), if ((-@z(y[8], 1@z) →* y[6])∧(z[8]* z[6])∧(x[8]* x[6]))


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


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


(9) -> (2), if ((y[9]* y[2])∧(z[9]* z[2])∧(x[9]* x[2]))


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


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


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


(9) -> (10), if ((y[9]* y[10])∧(z[9]* z[10])∧(x[9]* x[10]))


(10) -> (7), if ((z[10]* z[7])∧(x[10]* x[7])∧(y[10]* y[7])∧(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])) →* TRUE))


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


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


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


(11) -> (5), if ((-@z(y[11], 1@z) →* y[5])∧(z[11]* z[5])∧(x[11]* x[5]))


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


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




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




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




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




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




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




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




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




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




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




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




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




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL2(x1, x2, x3, x4)) = -1 + (-1)x4 + x1   
POL(COND_EVAL3(x1, x2, x3, x4)) = -1 + (-1)x4 + x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL5(x1, x2, x3, x4)) = -1 + (-1)x4 + x1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4   
POL(COND_EVAL4(x1, x2, x3, x4)) = -1 + (-1)x4   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL2(TRUE, x[9], y[9], z[9]) → EVAL(x[9], y[9], z[9])

The following pairs are in Pbound:

COND_EVAL2(TRUE, x[9], y[9], z[9]) → EVAL(x[9], y[9], z[9])

The following pairs are in P:

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

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

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


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

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


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


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


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


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


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


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


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


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


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


(8) -> (5), if ((-@z(y[8], 1@z) →* y[5])∧(z[8]* z[5])∧(x[8]* x[5]))


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


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


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


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


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


(10) -> (7), if ((z[10]* z[7])∧(x[10]* x[7])∧(y[10]* y[7])∧(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])) →* TRUE))


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


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


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


(8) -> (6), if ((-@z(y[8], 1@z) →* y[6])∧(z[8]* z[6])∧(x[8]* x[6]))


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


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


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


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


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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


(11) -> (5), if ((-@z(y[11], 1@z) →* y[5])∧(z[11]* z[5])∧(x[11]* x[5]))


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

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

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


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


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


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


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


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


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


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


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


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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


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


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


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


(10) -> (7), if ((z[10]* z[7])∧(x[10]* x[7])∧(y[10]* y[7])∧(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])) →* TRUE))


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


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


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


(8) -> (6), if ((-@z(y[8], 1@z) →* y[6])∧(z[8]* z[6])∧(x[8]* x[6]))


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


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


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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


For Pair COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11]) the following chains were created:




For Pair EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1]) the following chains were created:




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




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




For Pair COND_EVAL3(TRUE, x[8], y[8], z[8]) → EVAL(x[8], -@z(y[8], 1@z), z[8]) the following chains were created:




For Pair EVAL(x[3], y[3], z[3]) → COND_EVAL3(&&(&&(>@z(x[3], z[3]), >=@z(z[3], x[3])), >@z(y[3], z[3])), x[3], y[3], z[3]) the following chains were created:




For Pair COND_EVAL5(TRUE, x[7], y[7], z[7]) → EVAL(x[7], y[7], z[7]) the following chains were created:




For Pair EVAL(x[10], y[10], z[10]) → COND_EVAL5(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])), x[10], y[10], z[10]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4]) the following chains were created:




For Pair EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL3(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(TRUE) = 0   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + x2   
POL(FALSE) = 0   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL5(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(COND_EVAL4(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL5(TRUE, x[7], y[7], z[7]) → EVAL(x[7], y[7], z[7])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])
COND_EVAL3(TRUE, x[8], y[8], z[8]) → EVAL(x[8], -@z(y[8], 1@z), z[8])

The following pairs are in P:

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

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
            ↳ 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:

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

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


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


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


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


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


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


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


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


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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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


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


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


(10) -> (7), if ((z[10]* z[7])∧(x[10]* x[7])∧(y[10]* y[7])∧(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(7): COND_EVAL5(TRUE, x[7], y[7], z[7]) → EVAL(x[7], y[7], z[7])
(10): EVAL(x[10], y[10], z[10]) → COND_EVAL5(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])), x[10], y[10], z[10])
(4): COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])
(6): EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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


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


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


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


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


(10) -> (7), if ((z[10]* z[7])∧(x[10]* x[7])∧(y[10]* y[7])∧(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])) →* TRUE))


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


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




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




For Pair COND_EVAL5(TRUE, x[7], y[7], z[7]) → EVAL(x[7], y[7], z[7]) the following chains were created:




For Pair EVAL(x[10], y[10], z[10]) → COND_EVAL5(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])), x[10], y[10], z[10]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4]) the following chains were created:




For Pair EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL5(x1, x2, x3, x4)) = (-1)x4 + (-1)x3 + x2 + (2)x1   
POL(>=@z(x1, x2)) = -1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL4(x1, x2, x3, x4)) = -1 + (-1)x4 + (-1)x3 + x2   
POL(EVAL(x1, x2, x3)) = (-1)x3 + (-1)x2 + x1   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + (-1)x3 + 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[10], y[10], z[10]) → COND_EVAL5(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])), x[10], y[10], z[10])
EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

The following pairs are in Pbound:

COND_EVAL5(TRUE, x[7], y[7], z[7]) → EVAL(x[7], y[7], z[7])

The following pairs are in P:

COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
COND_EVAL5(TRUE, x[7], y[7], z[7]) → EVAL(x[7], y[7], z[7])
COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])

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

I DP problem:
The following domains are used:

z

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

(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(10): EVAL(x[10], y[10], z[10]) → COND_EVAL5(&&(&&(>@z(x[10], z[10]), >=@z(z[10], x[10])), >=@z(z[10], y[10])), x[10], y[10], z[10])
(4): COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])
(6): EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(4): COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])
(6): EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


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




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




For Pair COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4]) the following chains were created:




For Pair EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(COND_EVAL4(x1, x2, x3, x4)) = -1 + (-1)x4 + x2   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + x1   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + 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_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])

The following pairs are in Pbound:

COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])

The following pairs are in P:

COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

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

I DP problem:
The following domains are used:

z

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

(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(6): EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


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




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




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



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

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

The following pairs are in P>:

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

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])

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

I DP problem:
The following domains are used:

z

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

(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])


The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

I DP problem:
The following domains are used:

z

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

(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(7): COND_EVAL5(TRUE, x[7], y[7], z[7]) → EVAL(x[7], y[7], z[7])
(4): COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


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




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




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



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

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

The following pairs are in P>:

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

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])

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
            ↳ IDP
              ↳ IDependencyGraphProof
                ↳ 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:

(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])


The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

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

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


(8) -> (6), if ((-@z(y[8], 1@z) →* y[6])∧(z[8]* z[6])∧(x[8]* x[6]))


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


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


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


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


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


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


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


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


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


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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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


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


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


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


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

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

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


(8) -> (6), if ((-@z(y[8], 1@z) →* y[6])∧(z[8]* z[6])∧(x[8]* x[6]))


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


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


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


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


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


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


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


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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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


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


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


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


For Pair COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11]) the following chains were created:




For Pair EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1]) the following chains were created:




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




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




For Pair COND_EVAL3(TRUE, x[8], y[8], z[8]) → EVAL(x[8], -@z(y[8], 1@z), z[8]) the following chains were created:




For Pair EVAL(x[3], y[3], z[3]) → COND_EVAL3(&&(&&(>@z(x[3], z[3]), >=@z(z[3], x[3])), >@z(y[3], z[3])), x[3], y[3], z[3]) the following chains were created:




For Pair COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4]) the following chains were created:




For Pair EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL3(x1, x2, x3, x4)) = (-1)x4 + x3 + x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2, x3)) = (-1)x3 + x2   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + (-1)x1   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + (-1)x1   
POL(COND_EVAL4(x1, x2, x3, x4)) = (-1)x4 + x3   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL3(TRUE, x[8], y[8], z[8]) → EVAL(x[8], -@z(y[8], 1@z), z[8])

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[8], y[8], z[8]) → EVAL(x[8], -@z(y[8], 1@z), z[8])

The following pairs are in P:

COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])
EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])
COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
EVAL(x[3], y[3], z[3]) → COND_EVAL3(&&(&&(>@z(x[3], z[3]), >=@z(z[3], x[3])), >@z(y[3], z[3])), x[3], y[3], z[3])
COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])
EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

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
            ↳ 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:

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

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


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


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


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


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


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


(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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


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


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(11): COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])
(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(4): COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])
(6): EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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


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


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


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


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


For Pair COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11]) the following chains were created:




For Pair EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1]) the following chains were created:




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




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




For Pair COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4]) the following chains were created:




For Pair EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + x2 + (-1)x1   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(COND_EVAL4(x1, x2, x3, x4)) = -1 + (-1)x4 + x2   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + x2 + (-1)x1   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + x1   
POL(FALSE) = 2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])

The following pairs are in Pbound:

COND_EVAL4(TRUE, x[4], y[4], z[4]) → EVAL(-@z(x[4], 1@z), y[4], z[4])

The following pairs are in P:

COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])
EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])
COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

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

(11): COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])
(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(6): EVAL(x[6], y[6], z[6]) → COND_EVAL4(>@z(x[6], z[6]), x[6], y[6], z[6])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


(11) -> (6), if ((-@z(y[11], 1@z) →* y[6])∧(z[11]* z[6])∧(x[11]* x[6]))


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


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

(11): COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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


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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


For Pair COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11]) the following chains were created:




For Pair EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1]) the following chains were created:




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




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




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + (-1)x1   
POL(EVAL(x1, x2, x3)) = (-1)x3 + x2   
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[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])

The following pairs are in Pbound:

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

The following pairs are in P:

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

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

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


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

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

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(0): COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(-@z(x[0], 1@z), y[0], z[0])

(0) -> (2), if ((y[0]* y[2])∧(z[0]* z[2])∧(-@z(x[0], 1@z) →* x[2]))


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


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




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




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



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

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

The following pairs are in P>:

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

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])

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

(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])


The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

(11): COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], z[2]), >@z(x[2], z[2])), x[2], y[2], z[2])

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


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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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

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

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

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


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



The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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


For Pair EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11]) 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_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + x2   
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[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[11], y[11], z[11]) → EVAL(x[11], -@z(y[11], 1@z), z[11])

The following pairs are in P:

EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])

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

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


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

(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], z[1]), >=@z(z[1], x[1])), x[1], y[1], z[1])


The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval5(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
Cond_eval4(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval1(TRUE, x0, x1, x2)
Cond_eval3(TRUE, x0, x1, x2)


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