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_eval_1(TRUE, x, y, z) → eval_2(x, y, z)
Cond_eval_21(TRUE, x, y, z) → eval_1(x, +@z(y, 1@z), z)
eval_2(x, y, z) → Cond_eval_22(>@z(x, z), x, y, z)
Cond_eval_2(TRUE, x, y, z) → eval_1(-@z(x, 1@z), y, z)
eval_2(x, y, z) → Cond_eval_21(>@z(x, z), x, y, z)
Cond_eval_22(TRUE, x, y, z) → eval_1(x, y, +@z(z, 1@z))
eval_1(x, y, z) → Cond_eval_1(>@z(x, y), x, y, z)
eval_2(x, y, z) → Cond_eval_2(>=@z(z, x), x, y, z)

The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(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_eval_1(TRUE, x, y, z) → eval_2(x, y, z)
Cond_eval_21(TRUE, x, y, z) → eval_1(x, +@z(y, 1@z), z)
eval_2(x, y, z) → Cond_eval_22(>@z(x, z), x, y, z)
Cond_eval_2(TRUE, x, y, z) → eval_1(-@z(x, 1@z), y, z)
eval_2(x, y, z) → Cond_eval_21(>@z(x, z), x, y, z)
Cond_eval_22(TRUE, x, y, z) → eval_1(x, y, +@z(z, 1@z))
eval_1(x, y, z) → Cond_eval_1(>@z(x, y), x, y, z)
eval_2(x, y, z) → Cond_eval_2(>=@z(z, x), x, y, z)

The integer pair graph contains the following rules and edges:

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_2(>=@z(z[0], x[0]), x[0], y[0], z[0])
(1): COND_EVAL_2(TRUE, x[1], y[1], z[1]) → EVAL_1(-@z(x[1], 1@z), y[1], z[1])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(4): COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])
(7): EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])

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


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


(2) -> (3), if ((z[2]* z[3])∧(x[2]* x[3])∧(y[2]* y[3])∧(>@z(x[2], y[2]) →* TRUE))


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


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


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


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


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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))


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



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(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): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_2(>=@z(z[0], x[0]), x[0], y[0], z[0])
(1): COND_EVAL_2(TRUE, x[1], y[1], z[1]) → EVAL_1(-@z(x[1], 1@z), y[1], z[1])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(4): COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])
(7): EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])

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


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


(2) -> (3), if ((z[2]* z[3])∧(x[2]* x[3])∧(y[2]* y[3])∧(>@z(x[2], y[2]) →* TRUE))


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


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


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


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


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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))


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



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(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_2(x, y, z) → COND_EVAL_2(>=@z(z, x), x, y, z) the following chains were created:




For Pair COND_EVAL_2(TRUE, x, y, z) → EVAL_1(-@z(x, 1@z), y, z) the following chains were created:




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




For Pair COND_EVAL_1(TRUE, x, y, z) → EVAL_2(x, y, z) the following chains were created:




For Pair COND_EVAL_22(TRUE, x, y, z) → EVAL_1(x, y, +@z(z, 1@z)) the following chains were created:




For Pair COND_EVAL_21(TRUE, x, y, z) → EVAL_1(x, +@z(y, 1@z), z) the following chains were created:




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




For Pair EVAL_2(x, y, z) → COND_EVAL_22(>@z(x, z), x, 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_EVAL_1(x1, x2, x3, x4)) = -1 + (-1)x3 + x2   
POL(COND_EVAL_2(x1, x2, x3, x4)) = -1 + (-1)x3 + x2   
POL(TRUE) = 0   
POL(EVAL_2(x1, x2, x3)) = -1 + (-1)x2 + x1   
POL(COND_EVAL_21(x1, x2, x3, x4)) = -1 + (-1)x3 + x2   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL_1(x1, x2, x3)) = -1 + (-1)x2 + x1   
POL(COND_EVAL_22(x1, x2, x3, x4)) = -1 + (-1)x3 + x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL_2(TRUE, x[1], y[1], z[1]) → EVAL_1(-@z(x[1], 1@z), y[1], z[1])

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])

The following pairs are in P:

EVAL_2(x[0], y[0], z[0]) → COND_EVAL_2(>=@z(z[0], x[0]), x[0], y[0], z[0])
EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))
COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])
EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])

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

-@z1
+@z1


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

I DP problem:
The following domains are used:

z

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

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_2(>=@z(z[0], x[0]), x[0], y[0], z[0])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(4): COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])
(7): EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])

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


(2) -> (3), if ((z[2]* z[3])∧(x[2]* x[3])∧(y[2]* y[3])∧(>@z(x[2], y[2]) →* TRUE))


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


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


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


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


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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(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
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
IDP
                    ↳ IDPNonInfProof
              ↳ IDP

I DP problem:
The following domains are used:

z

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

(4): COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))
(7): EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])

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


(2) -> (3), if ((z[2]* z[3])∧(x[2]* x[3])∧(y[2]* y[3])∧(>@z(x[2], y[2]) →* TRUE))


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


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


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


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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(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_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z)) the following chains were created:




For Pair EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7]) the following chains were created:




For Pair COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3]) the following chains were created:




For Pair EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2]) the following chains were created:




For Pair COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5]) the following chains were created:




For Pair EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@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(EVAL_1(x1, x2, x3)) = -1 + (-1)x3 + x1   
POL(COND_EVAL_1(x1, x2, x3, x4)) = -1 + (-1)x4 + x2   
POL(TRUE) = -1   
POL(EVAL_2(x1, x2, x3)) = -1 + (-1)x3 + x1   
POL(COND_EVAL_22(x1, x2, x3, x4)) = -1 + (-1)x4 + x2   
POL(COND_EVAL_21(x1, x2, x3, x4)) = -1 + (-1)x4 + x2   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = 0   

The following pairs are in P>:

COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))

The following pairs are in Pbound:

COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))

The following pairs are in P:

EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])
COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])

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

+@z1


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

(7): EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])
(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])

(2) -> (3), if ((z[2]* z[3])∧(x[2]* x[3])∧(y[2]* y[3])∧(>@z(x[2], y[2]) →* TRUE))


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


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


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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(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
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
IDP
                            ↳ IDPNonInfProof
              ↳ IDP

I DP problem:
The following domains are used:

z

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

(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])

(2) -> (3), if ((z[2]* z[3])∧(x[2]* x[3])∧(y[2]* y[3])∧(>@z(x[2], y[2]) →* TRUE))


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


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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(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_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3]) the following chains were created:




For Pair EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2]) the following chains were created:




For Pair COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5]) the following chains were created:




For Pair EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@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(EVAL_1(x1, x2, x3)) = -1 + (-1)x2 + x1   
POL(COND_EVAL_1(x1, x2, x3, x4)) = -1 + (-1)x3 + x2   
POL(TRUE) = 2   
POL(EVAL_2(x1, x2, x3)) = -1 + (-1)x2 + x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND_EVAL_21(x1, x2, x3, x4)) = -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_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])

The following pairs are in P:

COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])

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

+@z1


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

I DP problem:
The following domains are used:

z

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

(3): COND_EVAL_1(TRUE, x[3], y[3], z[3]) → EVAL_2(x[3], y[3], z[3])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])

(2) -> (3), if ((z[2]* z[3])∧(x[2]* x[3])∧(y[2]* y[3])∧(>@z(x[2], y[2]) →* TRUE))


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



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

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

I DP problem:
The following domains are used:

z

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

(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])

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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.

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

I DP problem:
The following domains are used:

z

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

(0): EVAL_2(x[0], y[0], z[0]) → COND_EVAL_2(>=@z(z[0], x[0]), x[0], y[0], z[0])
(1): COND_EVAL_2(TRUE, x[1], y[1], z[1]) → EVAL_1(-@z(x[1], 1@z), y[1], z[1])
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_1(>@z(x[2], y[2]), x[2], y[2], z[2])
(4): COND_EVAL_22(TRUE, x[4], y[4], z[4]) → EVAL_1(x[4], y[4], +@z(z[4], 1@z))
(5): COND_EVAL_21(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], +@z(y[5], 1@z), z[5])
(6): EVAL_2(x[6], y[6], z[6]) → COND_EVAL_21(>@z(x[6], z[6]), x[6], y[6], z[6])
(7): EVAL_2(x[7], y[7], z[7]) → COND_EVAL_22(>@z(x[7], z[7]), x[7], y[7], z[7])

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


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


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


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


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


(6) -> (5), if ((z[6]* z[5])∧(x[6]* x[5])∧(y[6]* y[5])∧(>@z(x[6], z[6]) →* TRUE))



The set Q consists of the following terms:

Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_21(TRUE, x0, x1, x2)
eval_2(x0, x1, x2)
Cond_eval_2(TRUE, x0, x1, x2)
Cond_eval_22(TRUE, x0, x1, x2)
eval_1(x0, x1, x2)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 7 less nodes.