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:

exp(x, y) → if(>@z(y, 0@z), x, y)
cu(TRUE, x) → cu(<@z(x, exp(10@z, 2@z)), +@z(x, 1@z))
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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:

exp(x, y) → if(>@z(y, 0@z), x, y)
cu(TRUE, x) → cu(<@z(x, exp(10@z, 2@z)), +@z(x, 1@z))
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(0): CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))
(1): CU(TRUE, x[1]) → EXP(10@z, 2@z)
(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])
(3): IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))

(0) -> (0), if ((+@z(x[0], 1@z) →* x[0]a)∧(<@z(x[0], exp(10@z, 2@z)) →* TRUE))


(0) -> (1), if ((+@z(x[0], 1@z) →* x[1])∧(<@z(x[0], exp(10@z, 2@z)) →* TRUE))


(1) -> (2), if (EXP(10@z, 2@z) →* EXP(x[2], y[2]))


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


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



The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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
          ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(0): CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))
(1): CU(TRUE, x[1]) → EXP(10@z, 2@z)
(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])
(3): IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))

(0) -> (0), if ((+@z(x[0], 1@z) →* x[0]a)∧(<@z(x[0], exp(10@z, 2@z)) →* TRUE))


(0) -> (1), if ((+@z(x[0], 1@z) →* x[1])∧(<@z(x[0], exp(10@z, 2@z)) →* TRUE))


(1) -> (2), if (EXP(10@z, 2@z) →* EXP(x[2], y[2]))


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


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



The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(TRUE, x0, x1)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(3): IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))
(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])

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


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



The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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
          ↳ IDependencyGraphProof
            ↳ AND
              ↳ IDP
                ↳ UsableRulesProof
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): IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))
(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])

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


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



The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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 IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z)) the following chains were created:




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

The following pairs are in P>:

IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))

The following pairs are in Pbound:

IF(TRUE, x[3], y[3]) → EXP(x[3], -@z(y[3], 1@z))

The following pairs are in P:

EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])

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

-@z1


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

(2): EXP(x[2], y[2]) → IF(>@z(y[2], 0@z), x[2], y[2])


The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(TRUE, x0, x1)


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

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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph contains the following rules and edges:

(0): CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))

(0) -> (0), if ((+@z(x[0], 1@z) →* x[0]a)∧(<@z(x[0], exp(10@z, 2@z)) →* TRUE))



The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(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 CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 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(exp(x1, x2)) = -1 + (-1)x1   
POL(0@z) = 0   
POL(*@z(x1, x2)) = x1·x2   
POL(TRUE) = -1   
POL(2@z) = 2   
POL(FALSE) = -1   
POL(<@z(x1, x2)) = -1   
POL(>@z(x1, x2)) = -1   
POL(CU(x1, x2)) = -1 + (-1)x2   
POL(if(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(10@z) = 10   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))

The following pairs are in Pbound:

CU(TRUE, x[0]) → CU(<@z(x[0], exp(10@z, 2@z)), +@z(x[0], 1@z))

The following pairs are in P:
none

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

+@z1


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

I DP problem:
The following domains are used:

z

The ITRS R consists of the following rules:

exp(x, y) → if(>@z(y, 0@z), x, y)
if(FALSE, x, y) → 1@z
if(TRUE, x, y) → *@z(x, exp(x, -@z(y, 1@z)))

The integer pair graph is empty.
The set Q consists of the following terms:

exp(x0, x1)
cu(TRUE, x0)
if(FALSE, x0, x1)
if(TRUE, x0, x1)


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