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:

eval(x, y, z) → Cond_eval(&&(>=@z(x, z), >@z(y, 0@z)), x, y, z)
Cond_eval(TRUE, x, y, z) → eval(x, y, +@z(z, y))

The set Q consists of the following terms:

eval(x0, x1, x2)
Cond_eval(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:

eval(x, y, z) → Cond_eval(&&(>=@z(x, z), >@z(y, 0@z)), x, y, z)
Cond_eval(TRUE, x, y, z) → eval(x, y, +@z(z, y))

The integer pair graph contains the following rules and edges:

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

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


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



The set Q consists of the following terms:

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

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


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



The set Q consists of the following terms:

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




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

The following pairs are in P>:

COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(x[0], y[0], +@z(z[0], y[0]))

The following pairs are in Pbound:

COND_EVAL(TRUE, x[0], y[0], z[0]) → EVAL(x[0], y[0], +@z(z[0], y[0]))

The following pairs are in P:

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

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

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


↳ 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:

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


The set Q consists of the following terms:

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


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