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_eval1(TRUE, x, y, z) → eval(+@z(x, 1@z), y, z)
Cond_eval(TRUE, x, y, z) → eval(x, y, +@z(z, 1@z))
eval(x, y, z) → Cond_eval1(&&(>@z(y, x), >@z(z, x)), x, y, z)
eval(x, y, z) → Cond_eval(&&(>@z(y, x), >=@z(x, z)), x, y, z)

The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
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_eval1(TRUE, x, y, z) → eval(+@z(x, 1@z), y, z)
Cond_eval(TRUE, x, y, z) → eval(x, y, +@z(z, 1@z))
eval(x, y, z) → Cond_eval1(&&(>@z(y, x), >@z(z, x)), x, y, z)
eval(x, y, z) → Cond_eval(&&(>@z(y, x), >=@z(x, z)), x, y, z)

The integer pair graph contains the following rules and edges:

(0): COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], x[2]), >=@z(x[2], z[2])), x[2], y[2], z[2])
(3): COND_EVAL(TRUE, x[3], y[3], z[3]) → EVAL(x[3], y[3], +@z(z[3], 1@z))

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


(0) -> (2), if ((y[0] →* y[2])∧(z[0] →* z[2])∧(+@z(x[0], 1@z) →* x[2]))


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


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


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
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): COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], x[2]), >=@z(x[2], z[2])), x[2], y[2], z[2])
(3): COND_EVAL(TRUE, x[3], y[3], z[3]) → EVAL(x[3], y[3], +@z(z[3], 1@z))

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


(0) -> (2), if ((y[0] →* y[2])∧(z[0] →* z[2])∧(+@z(x[0], 1@z) →* x[2]))


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


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


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
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 COND_EVAL1(TRUE, x, y, z) → EVAL(+@z(x, 1@z), y, z) the following chains were created:




For Pair EVAL(x, y, z) → COND_EVAL1(&&(>@z(y, x), >@z(z, x)), x, y, z) the following chains were created:




For Pair EVAL(x, y, z) → COND_EVAL(&&(>@z(y, x), >=@z(x, z)), x, y, z) the following chains were created:




For Pair COND_EVAL(TRUE, x, y, z) → EVAL(x, y, +@z(z, 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)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4)) = -1 + (-1)x4 + (2)x3 + (-1)x2 + (-1)x1   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND_EVAL(x1, x2, x3, x4)) = -1 + (-1)x4 + (2)x3 + (-1)x2 + (-1)x1   
POL(EVAL(x1, x2, x3)) = -1 + (-1)x3 + (2)x2 + (-1)x1   
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[3], y[3], z[3]) → EVAL(x[3], y[3], +@z(z[3], 1@z))

The following pairs are in Pbound:

COND_EVAL(TRUE, x[3], y[3], z[3]) → EVAL(x[3], y[3], +@z(z[3], 1@z))

The following pairs are in P≥:

COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0])
EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1])
EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], x[2]), >=@z(x[2], z[2])), x[2], y[2], z[2])

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

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

(0): COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0])
(1): EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1])
(2): EVAL(x[2], y[2], z[2]) → COND_EVAL(&&(>@z(y[2], x[2]), >=@z(x[2], z[2])), x[2], y[2], z[2])

(0) -> (2), if ((y[0] →* y[2])∧(z[0] →* z[2])∧(+@z(x[0], 1@z) →* x[2]))


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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
Cond_eval(TRUE, x0, x1, x2)
eval(x0, x1, x2)


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

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

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_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1])
(0): COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0])

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


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



The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
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[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0]) 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(COND_EVAL1(x1, x2, x3, x4)) = -1 + x3 + (-1)x2 + (-1)x1   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(+@z(x1, x2)) = x1 + x2   
POL(EVAL(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(FALSE) = 0   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0])

The following pairs are in Pbound:

COND_EVAL1(TRUE, x[0], y[0], z[0]) → EVAL(+@z(x[0], 1@z), y[0], z[0])

The following pairs are in P≥:

EVAL(x[1], y[1], z[1]) → COND_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1])

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ IDP
              ↳ IDependencyGraphProof
                ↳ 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_EVAL1(&&(>@z(y[1], x[1]), >@z(z[1], x[1])), x[1], y[1], z[1])


The set Q consists of the following terms:

Cond_eval1(TRUE, x0, x1, x2)
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.