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

The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
eval_1(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(y, x)), x, y)
Cond_eval_1(TRUE, x, y) → eval_2(x, 1@z)
eval_2(x, y) → Cond_eval_2(&&(&&(>=@z(x, 0@z), >@z(y, 0@z)), >@z(x, y)), x, y)
Cond_eval_2(TRUE, x, y) → eval_2(x, *@z(2@z, y))
Cond_eval_21(TRUE, x, y) → eval_1(-@z(x, 1@z), y)
eval_1(x, y) → Cond_eval_1(>=@z(x, 0@z), x, 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_1(-@z(x[1], 1@z), y[1])
(2): COND_EVAL_1(TRUE, x[2], y[2]) → EVAL_2(x[2], 1@z)
(3): EVAL_2(x[3], y[3]) → COND_EVAL_21(&&(&&(>=@z(x[3], 0@z), >@z(y[3], 0@z)), >=@z(y[3], x[3])), x[3], y[3])
(4): COND_EVAL_2(TRUE, x[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4]))
(5): EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[5])

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


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


(2) -> (3), if ((x[2]* x[3]))


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


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


(4) -> (3), if ((*@z(2@z, y[4]) →* y[3])∧(x[4]* x[3]))


(4) -> (5), if ((*@z(2@z, y[4]) →* y[5])∧(x[4]* x[5]))


(5) -> (4), if ((x[5]* x[4])∧(y[5]* y[4])∧(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])) →* TRUE))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
eval_1(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_1(-@z(x[1], 1@z), y[1])
(2): COND_EVAL_1(TRUE, x[2], y[2]) → EVAL_2(x[2], 1@z)
(3): EVAL_2(x[3], y[3]) → COND_EVAL_21(&&(&&(>=@z(x[3], 0@z), >@z(y[3], 0@z)), >=@z(y[3], x[3])), x[3], y[3])
(4): COND_EVAL_2(TRUE, x[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4]))
(5): EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[5])

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


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


(2) -> (3), if ((x[2]* x[3]))


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


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


(4) -> (3), if ((*@z(2@z, y[4]) →* y[3])∧(x[4]* x[3]))


(4) -> (5), if ((*@z(2@z, y[4]) →* y[5])∧(x[4]* x[5]))


(5) -> (4), if ((x[5]* x[4])∧(y[5]* y[4])∧(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])) →* TRUE))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
eval_1(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_1(-@z(x, 1@z), y) the following chains were created:




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




For Pair COND_EVAL_2(TRUE, x, y) → EVAL_2(x, *@z(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(x, y)), x, y) 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(*@z(x1, x2)) = x1·x2   
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)) = x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL_1(TRUE, x[2], y[2]) → EVAL_2(x[2], 1@z)

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[2], y[2]) → EVAL_2(x[2], 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_1(-@z(x[1], 1@z), y[1])
EVAL_2(x[3], y[3]) → COND_EVAL_21(&&(&&(>=@z(x[3], 0@z), >@z(y[3], 0@z)), >=@z(y[3], x[3])), x[3], y[3])
COND_EVAL_2(TRUE, x[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4]))
EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[5])

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

*@z1
&&(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): 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_1(-@z(x[1], 1@z), y[1])
(3): EVAL_2(x[3], y[3]) → COND_EVAL_21(&&(&&(>=@z(x[3], 0@z), >@z(y[3], 0@z)), >=@z(y[3], x[3])), x[3], y[3])
(4): COND_EVAL_2(TRUE, x[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4]))
(5): EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[5])

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


(5) -> (4), if ((x[5]* x[4])∧(y[5]* y[4])∧(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])) →* TRUE))


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


(4) -> (5), if ((*@z(2@z, y[4]) →* y[5])∧(x[4]* x[5]))


(4) -> (3), if ((*@z(2@z, y[4]) →* y[3])∧(x[4]* x[3]))



The set Q consists of the following terms:

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


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

↳ 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:

(4): COND_EVAL_2(TRUE, x[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4]))
(5): EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[5])

(5) -> (4), if ((x[5]* x[4])∧(y[5]* y[4])∧(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])) →* TRUE))


(4) -> (5), if ((*@z(2@z, y[4]) →* y[5])∧(x[4]* x[5]))



The set Q consists of the following terms:

eval_2(x0, x1)
Cond_eval_1(TRUE, x0, x1)
Cond_eval_2(TRUE, x0, x1)
Cond_eval_21(TRUE, x0, x1)
eval_1(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[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4])) the following chains were created:




For Pair EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[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)) = -1   
POL(0@z) = 0   
POL(*@z(x1, x2)) = x1·x2   
POL(EVAL_2(x1, x2)) = 1 + (-1)x2 + x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL_2(x1, x2, x3)) = (-1)x3 + x2 + (-1)x1   
POL(2@z) = 2   
POL(FALSE) = -1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL_2(TRUE, x[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4]))

The following pairs are in Pbound:

COND_EVAL_2(TRUE, x[4], y[4]) → EVAL_2(x[4], *@z(2@z, y[4]))

The following pairs are in P:

EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[5])

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(5): EVAL_2(x[5], y[5]) → COND_EVAL_2(&&(&&(>=@z(x[5], 0@z), >@z(y[5], 0@z)), >@z(x[5], y[5])), x[5], y[5])


The set Q consists of the following terms:

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


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