Termination of the given ITRSProblem could successfully be proven:



ITRS
  ↳ ITRStoIDPProof

ITRS problem:
The following domains are used:

z

The TRS R consists of the following rules:

eval_2(x, y) → Cond_eval_21(&&(&&(>=@z(x, 0@z), >@z(y, 0@z)), >=@z(x, y)), x, y)
Cond_eval_1(TRUE, x, y) → eval_2(+@z(x, 1@z), 1@z)
eval_2(x, y) → Cond_eval_2(&&(&&(>=@z(x, 0@z), >@z(y, 0@z)), >@z(y, x)), x, y)
eval_1(x, y) → Cond_eval_1(>=@z(x, 0@z), x, y)
Cond_eval_21(TRUE, x, y) → eval_2(x, +@z(y, 1@z))
Cond_eval_2(TRUE, x, y) → eval_1(-@z(x, 2@z), y)

The set Q consists of the following terms:

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


Added dependency pairs

↳ ITRS
  ↳ ITRStoIDPProof
IDP
      ↳ UsableRulesProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

eval_2(x, y) → Cond_eval_21(&&(&&(>=@z(x, 0@z), >@z(y, 0@z)), >=@z(x, y)), x, y)
Cond_eval_1(TRUE, x, y) → eval_2(+@z(x, 1@z), 1@z)
eval_2(x, y) → Cond_eval_2(&&(&&(>=@z(x, 0@z), >@z(y, 0@z)), >@z(y, x)), x, y)
eval_1(x, y) → Cond_eval_1(>=@z(x, 0@z), x, y)
Cond_eval_21(TRUE, x, y) → eval_2(x, +@z(y, 1@z))
Cond_eval_2(TRUE, x, y) → eval_1(-@z(x, 2@z), y)

The integer pair graph contains the following rules and edges:

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

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


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


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


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


(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], 2@z) →* x[0]))


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


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


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



The set Q consists of the following terms:

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


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
IDP
          ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

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

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

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


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


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


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


(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], 2@z) →* x[0]))


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


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


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



The set Q consists of the following terms:

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


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


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




For Pair COND_EVAL_21(TRUE, x, y) → EVAL_2(x, +@z(y, 1@z)) the following chains were created:




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




For Pair COND_EVAL_2(TRUE, x, y) → EVAL_1(-@z(x, 2@z), y) the following chains were created:




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




For Pair COND_EVAL_1(TRUE, x, y) → EVAL_2(+@z(x, 1@z), 1@z) the following chains were created:




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL_1(x1, x2)) = x1   
POL(2@z) = 2   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(EVAL_2(x1, x2)) = -1 + x1   
POL(COND_EVAL_1(x1, x2, x3)) = x2   
POL(COND_EVAL_2(x1, x2, x3)) = -1 + x2   
POL(COND_EVAL_21(x1, x2, x3)) = -1 + 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[3], y[3]) → EVAL_1(-@z(x[3], 2@z), y[3])

The following pairs are in Pbound:

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

The following pairs are in P:

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

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

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


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

I DP problem:
The following domains are used:

z

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

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

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


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


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


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


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


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



The set Q consists of the following terms:

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


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
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): COND_EVAL_21(TRUE, x[1], y[1]) → EVAL_2(x[1], +@z(y[1], 1@z))
(2): EVAL_2(x[2], y[2]) → COND_EVAL_21(&&(&&(>=@z(x[2], 0@z), >@z(y[2], 0@z)), >=@z(x[2], y[2])), x[2], y[2])

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


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



The set Q consists of the following terms:

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


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


For Pair COND_EVAL_21(TRUE, x[1], y[1]) → EVAL_2(x[1], +@z(y[1], 1@z)) the following chains were created:




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

The following pairs are in P>:

EVAL_2(x[2], y[2]) → COND_EVAL_21(&&(&&(>=@z(x[2], 0@z), >@z(y[2], 0@z)), >=@z(x[2], y[2])), x[2], y[2])

The following pairs are in Pbound:

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

The following pairs are in P:

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

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

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


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

I DP problem:
The following domains are used:

z

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

(2): EVAL_2(x[2], y[2]) → COND_EVAL_21(&&(&&(>=@z(x[2], 0@z), >@z(y[2], 0@z)), >=@z(x[2], y[2])), x[2], y[2])


The set Q consists of the following terms:

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


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

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

(1): COND_EVAL_21(TRUE, x[1], y[1]) → EVAL_2(x[1], +@z(y[1], 1@z))


The set Q consists of the following terms:

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


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

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

I DP problem:
The following domains are used:

z

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

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

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


(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], 2@z) →* x[0]))


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


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


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



The set Q consists of the following terms:

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


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

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

I DP problem:
The following domains are used:

z

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

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

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


(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], 2@z) →* x[0]))


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


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



The set Q consists of the following terms:

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


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


For Pair COND_EVAL_2(TRUE, x[3], y[3]) → EVAL_1(-@z(x[3], 2@z), y[3]) the following chains were created:




For Pair EVAL_2(x[4], y[4]) → COND_EVAL_2(&&(&&(>=@z(x[4], 0@z), >@z(y[4], 0@z)), >@z(y[4], x[4])), x[4], y[4]) the following chains were created:




For Pair COND_EVAL_1(TRUE, x[5], y[5]) → EVAL_2(+@z(x[5], 1@z), 1@z) the following chains were created:




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




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



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

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(0@z) = 0   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(EVAL_1(x1, x2)) = 1 + x1   
POL(2@z) = 2   
POL(FALSE) = 1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL_1(x1, x2, x3)) = 1 + x2   
POL(EVAL_2(x1, x2)) = x1   
POL(COND_EVAL_2(x1, x2, x3)) = 1 + x2 + (-1)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL_2(TRUE, x[3], y[3]) → EVAL_1(-@z(x[3], 2@z), y[3])

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[5], y[5]) → EVAL_2(+@z(x[5], 1@z), 1@z)

The following pairs are in P:

EVAL_2(x[4], y[4]) → COND_EVAL_2(&&(&&(>=@z(x[4], 0@z), >@z(y[4], 0@z)), >@z(y[4], x[4])), x[4], y[4])
COND_EVAL_1(TRUE, x[5], y[5]) → EVAL_2(+@z(x[5], 1@z), 1@z)
EVAL_1(x[0], y[0]) → COND_EVAL_1(>=@z(x[0], 0@z), x[0], y[0])

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

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


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

I DP problem:
The following domains are used:

z

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

(4): EVAL_2(x[4], y[4]) → COND_EVAL_2(&&(&&(>=@z(x[4], 0@z), >@z(y[4], 0@z)), >@z(y[4], x[4])), x[4], y[4])
(5): COND_EVAL_1(TRUE, x[5], y[5]) → EVAL_2(+@z(x[5], 1@z), 1@z)
(0): EVAL_1(x[0], y[0]) → COND_EVAL_1(>=@z(x[0], 0@z), x[0], y[0])

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


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



The set Q consists of the following terms:

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


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
                  ↳ 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:

(3): COND_EVAL_2(TRUE, x[3], y[3]) → EVAL_1(-@z(x[3], 2@z), y[3])
(4): EVAL_2(x[4], y[4]) → COND_EVAL_2(&&(&&(>=@z(x[4], 0@z), >@z(y[4], 0@z)), >@z(y[4], x[4])), x[4], y[4])
(0): EVAL_1(x[0], y[0]) → COND_EVAL_1(>=@z(x[0], 0@z), x[0], y[0])

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


(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], 2@z) →* x[0]))



The set Q consists of the following terms:

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


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