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, y, z) → eval(-@z(x, 1@z), -@z(y, 1@z), z)
eval(x, y, z) → Cond_eval(&&(>@z(x, z), >@z(y, z)), x, y, z)

The set Q consists of the following terms:

Cond_eval(TRUE, x0, x1, x2)
eval(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:

Cond_eval(TRUE, x, y, z) → eval(-@z(x, 1@z), -@z(y, 1@z), z)
eval(x, y, z) → Cond_eval(&&(>@z(x, z), >@z(y, z)), x, y, z)

The integer pair graph contains the following rules and edges:

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

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


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



The set Q consists of the following terms:

Cond_eval(TRUE, x0, x1, x2)
eval(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): EVAL(x[0], y[0], z[0]) → COND_EVAL(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])
(1): COND_EVAL(TRUE, x[1], y[1], z[1]) → EVAL(-@z(x[1], 1@z), -@z(y[1], 1@z), z[1])

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


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



The set Q consists of the following terms:

Cond_eval(TRUE, x0, x1, x2)
eval(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 EVAL(x, y, z) → COND_EVAL(&&(>@z(x, z), >@z(y, z)), x, y, z) the following chains were created:




For Pair COND_EVAL(TRUE, x, y, z) → EVAL(-@z(x, 1@z), -@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(-@z(x1, x2)) = x1 + (-1)x2   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + x3   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

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

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[0], y[0], z[0]) → COND_EVAL(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])

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

&&(FALSE, FALSE)1FALSE1
-@z1
&&(TRUE, TRUE)1TRUE1
&&(FALSE, TRUE)1FALSE1
&&(TRUE, FALSE)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(x[0], y[0], z[0]) → COND_EVAL(&&(>@z(x[0], z[0]), >@z(y[0], z[0])), x[0], y[0], z[0])


The set Q consists of the following terms:

Cond_eval(TRUE, x0, x1, x2)
eval(x0, x1, x2)


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