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:

f(x, y, z) → Cond_f1(&&(>@z(x, y), >@z(x, z)), x, y, z)
f(x, y, z) → Cond_f(&&(>@z(x, y), >@z(x, z)), x, y, z)
Cond_f(TRUE, x, y, z) → f(x, +@z(y, 1@z), z)
Cond_f1(TRUE, x, y, z) → f(x, y, +@z(z, 1@z))

The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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:

f(x, y, z) → Cond_f1(&&(>@z(x, y), >@z(x, z)), x, y, z)
f(x, y, z) → Cond_f(&&(>@z(x, y), >@z(x, z)), x, y, z)
Cond_f(TRUE, x, y, z) → f(x, +@z(y, 1@z), z)
Cond_f1(TRUE, x, y, z) → f(x, y, +@z(z, 1@z))

The integer pair graph contains the following rules and edges:

(0): F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])
(1): COND_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z))
(2): F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(3): COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

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


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


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


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


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


(3) -> (2), if ((+@z(y[3], 1@z) →* y[2])∧(z[3] →* z[2])∧(x[3] →* x[2]))



The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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): F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])
(1): COND_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z))
(2): F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(3): COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

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


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


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


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


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


(3) -> (2), if ((+@z(y[3], 1@z) →* y[2])∧(z[3] →* z[2])∧(x[3] →* x[2]))



The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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 F(x, y, z) → COND_F1(&&(>@z(x, y), >@z(x, z)), x, y, z) the following chains were created:




For Pair COND_F1(TRUE, x, y, z) → F(x, y, +@z(z, 1@z)) the following chains were created:




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




For Pair COND_F(TRUE, x, y, z) → F(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(COND_F(x1, x2, x3, x4)) = -1 + (-1)x4 + (-1)x3 + (2)x2 + (-1)x1   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 0   
POL(COND_F1(x1, x2, x3, x4)) = (-1)x4 + (-1)x3 + (2)x2 + x1   
POL(F(x1, x2, x3)) = (-1)x3 + (-1)x2 + (2)x1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z))

The following pairs are in Pbound:

COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

The following pairs are in P≥:

F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])
F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

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

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


↳ 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): F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])
(2): F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(3): COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

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


(3) -> (2), if ((+@z(y[3], 1@z) →* y[2])∧(z[3] →* z[2])∧(x[3] →* x[2]))


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



The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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
            ↳ 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): F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])
(3): COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

(3) -> (2), if ((+@z(y[3], 1@z) →* y[2])∧(z[3] →* z[2])∧(x[3] →* x[2]))


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



The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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 F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2]) the following chains were created:




For Pair COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), 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(COND_F(x1, x2, x3, x4)) = -1 + (-1)x3 + x2 + (-1)x1   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 1   
POL(F(x1, x2, x3)) = -1 + (-1)x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])

The following pairs are in Pbound:

COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

The following pairs are in P≥:

COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])

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

&&(FALSE, FALSE)1 ↔ FALSE1
+@z1 ↔
&&(TRUE, TRUE)1 ↔ TRUE1
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): F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])


The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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
            ↳ 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:

(3): COND_F(TRUE, x[3], y[3], z[3]) → F(x[3], +@z(y[3], 1@z), z[3])


The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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
            ↳ 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): F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])
(1): COND_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z))
(2): F(x[2], y[2], z[2]) → COND_F(&&(>@z(x[2], y[2]), >@z(x[2], z[2])), x[2], y[2], z[2])

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


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


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



The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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
            ↳ 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): COND_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z))
(0): F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])

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


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



The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(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_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z)) the following chains were created:




For Pair F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], 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(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(COND_F1(x1, x2, x3, x4)) = -1 + (-1)x4 + (-1)x3 + (2)x2 + (-1)x1   
POL(F(x1, x2, x3)) = (-1)x3 + (-1)x2 + (2)x1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z))

The following pairs are in Pbound:

COND_F1(TRUE, x[1], y[1], z[1]) → F(x[1], y[1], +@z(z[1], 1@z))

The following pairs are in P≥:

F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(0): F(x[0], y[0], z[0]) → COND_F1(&&(>@z(x[0], y[0]), >@z(x[0], z[0])), x[0], y[0], z[0])


The set Q consists of the following terms:

f(x0, x1, x2)
Cond_f(TRUE, x0, x1, x2)
Cond_f1(TRUE, x0, x1, x2)


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