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_f(TRUE, x) → f(x)
f(x) → Cond_f(>@z(x, x), x)

The set Q consists of the following terms:

Cond_f(TRUE, x0)
f(x0)


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_f(TRUE, x) → f(x)
f(x) → Cond_f(>@z(x, x), x)

The integer pair graph contains the following rules and edges:

(0): COND_F(TRUE, x[0]) → F(x[0])
(1): F(x[1]) → COND_F(>@z(x[1], x[1]), x[1])

(0) -> (1), if ((x[0]* x[1]))


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



The set Q consists of the following terms:

Cond_f(TRUE, x0)
f(x0)


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): COND_F(TRUE, x[0]) → F(x[0])
(1): F(x[1]) → COND_F(>@z(x[1], x[1]), x[1])

(0) -> (1), if ((x[0]* x[1]))


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



The set Q consists of the following terms:

Cond_f(TRUE, x0)
f(x0)


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_F(TRUE, x) → F(x) the following chains were created:




For Pair F(x) → COND_F(>@z(x, x), x) 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(FALSE) = -1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   
POL(F(x1)) = 0   
POL(COND_F(x1, x2)) = -1   

The following pairs are in P>:

COND_F(TRUE, x[0]) → F(x[0])
F(x[1]) → COND_F(>@z(x[1], x[1]), x[1])

The following pairs are in Pbound:

COND_F(TRUE, x[0]) → F(x[0])
F(x[1]) → COND_F(>@z(x[1], x[1]), x[1])

The following pairs are in P:
none

There are no usable rules.

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

I DP problem:
The following domains are used:none

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

Cond_f(TRUE, x0)
f(x0)


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