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

The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(x0, x1, x2)


Added dependency pairs

↳ ITRS
  ↳ ITRStoIDPProof
IDP
      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

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

The integer pair graph contains the following rules and edges:

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

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


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


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


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


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


(3) -> (4), if ((z[3]* z[4])∧(x[3]* x[4])∧(y[3]* y[4])∧(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)) →* TRUE))


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


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


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


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


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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(x[0], y[0], z[0])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(&&(>@z(+@z(x[1], y[1]), z[1]), >=@z(z[1], 0@z)), >@z(x[1], 0@z)), x[1], y[1], z[1])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(&&(&&(>@z(+@z(x[2], y[2]), z[2]), >=@z(z[2], 0@z)), >=@z(0@z, x[2])), >=@z(0@z, y[2])), x[2], y[2], z[2])
(3): EVAL(x[3], y[3], z[3]) → COND_EVAL2(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])
(4): COND_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4])
(5): COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5])

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


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


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


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


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


(3) -> (4), if ((z[3]* z[4])∧(x[3]* x[4])∧(y[3]* y[4])∧(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)) →* TRUE))


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


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


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


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


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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(x, y, z) the following chains were created:




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




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




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




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




For Pair COND_EVAL1(TRUE, x, y, z) → EVAL(-@z(x, 1@z), y, 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 + (-1)x1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2, x3)) = (-1)x3   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + (-1)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + (-1)x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

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

The following pairs are in Pbound:

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

The following pairs are in P:

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

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

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

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

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


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


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


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


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


(3) -> (4), if ((z[3]* z[4])∧(x[3]* x[4])∧(y[3]* y[4])∧(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)) →* TRUE))


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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:

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

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


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


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


(3) -> (4), if ((z[3]* z[4])∧(x[3]* x[4])∧(y[3]* y[4])∧(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)) →* TRUE))


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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(+@z(x[1], y[1]), z[1]), >=@z(z[1], 0@z)), >@z(x[1], 0@z)), x[1], y[1], z[1]) the following chains were created:




For Pair COND_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4]) the following chains were created:




For Pair EVAL(x[3], y[3], z[3]) → COND_EVAL2(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)), x[3], y[3], z[3]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5]) 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 + x3 + x2 + (-1)x1   
POL(0@z) = 0   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + x2 + x1   
POL(FALSE) = 1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + x2 + (-1)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

EVAL(x[3], y[3], z[3]) → COND_EVAL2(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5])

The following pairs are in P:

EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(&&(>@z(+@z(x[1], y[1]), z[1]), >=@z(z[1], 0@z)), >@z(x[1], 0@z)), x[1], y[1], z[1])
COND_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4])
COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5])

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

FALSE1&&(FALSE, FALSE)1
-@z1
TRUE1&&(TRUE, TRUE)1
&&(FALSE, TRUE)1FALSE1
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(+@z(x[1], y[1]), z[1]), >=@z(z[1], 0@z)), >@z(x[1], 0@z)), x[1], y[1], z[1])
(4): COND_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4])
(5): COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5])

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


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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
                      ↳ 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(+@z(x[1], y[1]), z[1]), >=@z(z[1], 0@z)), >@z(x[1], 0@z)), x[1], y[1], z[1])
(5): COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5])

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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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(+@z(x[1], y[1]), z[1]), >=@z(z[1], 0@z)), >@z(x[1], 0@z)), x[1], y[1], z[1]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(>=@z(x1, x2)) = -1   
POL(0@z) = 0   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + (2)x2 + (-1)x1   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(+@z(x1, x2)) = x1 + x2   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + x2 + (2)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[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[5], y[5], z[5]) → EVAL(-@z(x[5], 1@z), y[5], z[5])

The following pairs are in P:

EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(&&(>@z(+@z(x[1], y[1]), z[1]), >=@z(z[1], 0@z)), >@z(x[1], 0@z)), 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
&&(TRUE, FALSE)1FALSE1
&&(FALSE, TRUE)1FALSE1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

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


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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:

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

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


(3) -> (4), if ((z[3]* z[4])∧(x[3]* x[4])∧(y[3]* y[4])∧(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)) →* TRUE))


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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:

(4): COND_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4])
(3): EVAL(x[3], y[3], z[3]) → COND_EVAL2(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])

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


(3) -> (4), if ((z[3]* z[4])∧(x[3]* x[4])∧(y[3]* y[4])∧(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)) →* TRUE))



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(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_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4]) the following chains were created:




For Pair EVAL(x[3], y[3], z[3]) → COND_EVAL2(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)), x[3], y[3], z[3]) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(COND_EVAL2(x1, x2, x3, x4)) = -1 + (-1)x4 + x3 + x2 + (-1)x1   
POL(>=@z(x1, x2)) = -1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(EVAL(x1, x2, x3)) = (-1)x3 + x2 + x1   
POL(FALSE) = 1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4])

The following pairs are in Pbound:

COND_EVAL2(TRUE, x[4], y[4], z[4]) → EVAL(x[4], -@z(y[4], 1@z), z[4])

The following pairs are in P:

EVAL(x[3], y[3], z[3]) → COND_EVAL2(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])

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

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], z[3]) → COND_EVAL2(&&(&&(&&(>@z(+@z(x[3], y[3]), z[3]), >=@z(z[3], 0@z)), >=@z(0@z, x[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
Cond_eval2(TRUE, x0, x1, x2)
eval(x0, x1, x2)


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