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_eval(TRUE, x) → eval(-@z(x, 1@z))
eval(x) → Cond_eval(&&(=@z(%@z(x, 2@z), 0@z), >@z(x, 0@z)), x)

The set Q consists of the following terms:

Cond_eval(TRUE, x0)
eval(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_eval(TRUE, x) → eval(-@z(x, 1@z))
eval(x) → Cond_eval(&&(=@z(%@z(x, 2@z), 0@z), >@z(x, 0@z)), x)

The integer pair graph contains the following rules and edges:

(0): EVAL(x[0]) → COND_EVAL(&&(=@z(%@z(x[0], 2@z), 0@z), >@z(x[0], 0@z)), x[0])
(1): COND_EVAL(TRUE, x[1]) → EVAL(-@z(x[1], 1@z))

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


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



The set Q consists of the following terms:

Cond_eval(TRUE, x0)
eval(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): EVAL(x[0]) → COND_EVAL(&&(=@z(%@z(x[0], 2@z), 0@z), >@z(x[0], 0@z)), x[0])
(1): COND_EVAL(TRUE, x[1]) → EVAL(-@z(x[1], 1@z))

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


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



The set Q consists of the following terms:

Cond_eval(TRUE, x0)
eval(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 EVAL(x) → COND_EVAL(&&(=@z(%@z(x, 2@z), 0@z), >@z(x, 0@z)), x) the following chains were created:




For Pair COND_EVAL(TRUE, x) → EVAL(-@z(x, 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(=@z(x1, x2)) = -1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = 2   
POL(2@z) = 2   
POL(COND_EVAL(x1, x2)) = -1 + (2)x2   
POL(EVAL(x1)) = (2)x1   
POL(FALSE) = 2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

Polynomial Interpretations with Context Sensitive Arithemetic Replacement
POL(TermCSAR-Mode @ Context)

POL(%@z(x1, 2@z)1 @ {}) = max{x2, (-1)x2}   
POL(%@z(x1, 2@z)-1 @ {}) = min{x2, (-1)x2}   

The following pairs are in P>:

EVAL(x[0]) → COND_EVAL(&&(=@z(%@z(x[0], 2@z), 0@z), >@z(x[0], 0@z)), x[0])
COND_EVAL(TRUE, x[1]) → EVAL(-@z(x[1], 1@z))

The following pairs are in Pbound:

COND_EVAL(TRUE, x[1]) → EVAL(-@z(x[1], 1@z))

The following pairs are in P:
none

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

&&(FALSE, FALSE)1FALSE1
-@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(x[0]) → COND_EVAL(&&(=@z(%@z(x[0], 2@z), 0@z), >@z(x[0], 0@z)), x[0])


The set Q consists of the following terms:

Cond_eval(TRUE, x0)
eval(x0)


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:none

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

Cond_eval(TRUE, x0)
eval(x0)


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