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_1(x, y, z) → Cond_eval_1(&&(&&(>@z(y, x), >@z(z, y)), >@z(y, 0@z)), x, y, z)
Cond_eval_11(TRUE, x, y, z) → eval_1(x, y, -@z(x, y))
Cond_eval_1(TRUE, x, y, z) → eval_1(+@z(x, y), y, z)
Cond_eval_0(TRUE, x, y, z) → eval_1(x, y, z)
eval_0(x, y, z) → Cond_eval_0(>@z(y, 0@z), x, y, z)
eval_1(x, y, z) → Cond_eval_11(&&(&&(>@z(y, x), >@z(z, y)), >@z(y, 0@z)), x, y, z)

The set Q consists of the following terms:

eval_1(x0, x1, x2)
Cond_eval_11(TRUE, x0, x1, x2)
Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_0(TRUE, x0, x1, x2)
eval_0(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_1(x, y, z) → Cond_eval_1(&&(&&(>@z(y, x), >@z(z, y)), >@z(y, 0@z)), x, y, z)
Cond_eval_11(TRUE, x, y, z) → eval_1(x, y, -@z(x, y))
Cond_eval_1(TRUE, x, y, z) → eval_1(+@z(x, y), y, z)
Cond_eval_0(TRUE, x, y, z) → eval_1(x, y, z)
eval_0(x, y, z) → Cond_eval_0(>@z(y, 0@z), x, y, z)
eval_1(x, y, z) → Cond_eval_11(&&(&&(>@z(y, x), >@z(z, y)), >@z(y, 0@z)), x, y, z)

The integer pair graph contains the following rules and edges:

(0): COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])
(1): COND_EVAL_11(TRUE, x[1], y[1], z[1]) → EVAL_1(x[1], y[1], -@z(x[1], y[1]))
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_11(&&(&&(>@z(y[2], x[2]), >@z(z[2], y[2])), >@z(y[2], 0@z)), x[2], y[2], z[2])
(3): EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])
(4): EVAL_0(x[4], y[4], z[4]) → COND_EVAL_0(>@z(y[4], 0@z), x[4], y[4], z[4])
(5): COND_EVAL_0(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], y[5], z[5])

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


(0) -> (3), if ((y[0]* y[3])∧(z[0]* z[3])∧(+@z(x[0], y[0]) →* x[3]))


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


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


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


(3) -> (0), if ((z[3]* z[0])∧(x[3]* x[0])∧(y[3]* y[0])∧(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)) →* TRUE))


(4) -> (5), if ((z[4]* z[5])∧(x[4]* x[5])∧(y[4]* y[5])∧(>@z(y[4], 0@z) →* TRUE))


(5) -> (2), if ((y[5]* y[2])∧(z[5]* z[2])∧(x[5]* x[2]))


(5) -> (3), if ((y[5]* y[3])∧(z[5]* z[3])∧(x[5]* x[3]))



The set Q consists of the following terms:

eval_1(x0, x1, x2)
Cond_eval_11(TRUE, x0, x1, x2)
Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_0(TRUE, x0, x1, x2)
eval_0(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
          ↳ 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_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])
(1): COND_EVAL_11(TRUE, x[1], y[1], z[1]) → EVAL_1(x[1], y[1], -@z(x[1], y[1]))
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_11(&&(&&(>@z(y[2], x[2]), >@z(z[2], y[2])), >@z(y[2], 0@z)), x[2], y[2], z[2])
(3): EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])
(4): EVAL_0(x[4], y[4], z[4]) → COND_EVAL_0(>@z(y[4], 0@z), x[4], y[4], z[4])
(5): COND_EVAL_0(TRUE, x[5], y[5], z[5]) → EVAL_1(x[5], y[5], z[5])

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


(0) -> (3), if ((y[0]* y[3])∧(z[0]* z[3])∧(+@z(x[0], y[0]) →* x[3]))


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


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


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


(3) -> (0), if ((z[3]* z[0])∧(x[3]* x[0])∧(y[3]* y[0])∧(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)) →* TRUE))


(4) -> (5), if ((z[4]* z[5])∧(x[4]* x[5])∧(y[4]* y[5])∧(>@z(y[4], 0@z) →* TRUE))


(5) -> (2), if ((y[5]* y[2])∧(z[5]* z[2])∧(x[5]* x[2]))


(5) -> (3), if ((y[5]* y[3])∧(z[5]* z[3])∧(x[5]* x[3]))



The set Q consists of the following terms:

eval_1(x0, x1, x2)
Cond_eval_11(TRUE, x0, x1, x2)
Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_0(TRUE, x0, x1, x2)
eval_0(x0, x1, x2)


The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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): COND_EVAL_11(TRUE, x[1], y[1], z[1]) → EVAL_1(x[1], y[1], -@z(x[1], y[1]))
(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_11(&&(&&(>@z(y[2], x[2]), >@z(z[2], y[2])), >@z(y[2], 0@z)), x[2], y[2], z[2])
(3): EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])
(0): COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])

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


(3) -> (0), if ((z[3]* z[0])∧(x[3]* x[0])∧(y[3]* y[0])∧(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)) →* TRUE))


(0) -> (3), if ((y[0]* y[3])∧(z[0]* z[3])∧(+@z(x[0], y[0]) →* x[3]))


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


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


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



The set Q consists of the following terms:

eval_1(x0, x1, x2)
Cond_eval_11(TRUE, x0, x1, x2)
Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_0(TRUE, x0, x1, x2)
eval_0(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_11(TRUE, x[1], y[1], z[1]) → EVAL_1(x[1], y[1], -@z(x[1], y[1])) the following chains were created:




For Pair EVAL_1(x[2], y[2], z[2]) → COND_EVAL_11(&&(&&(>@z(y[2], x[2]), >@z(z[2], y[2])), >@z(y[2], 0@z)), x[2], y[2], z[2]) the following chains were created:




For Pair EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3]) the following chains were created:




For Pair COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), 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(-@z(x1, x2)) = x1 + (-1)x2   
POL(0@z) = 0   
POL(EVAL_1(x1, x2, x3)) = -1 + x3 + (-1)x1   
POL(COND_EVAL_1(x1, x2, x3, x4)) = -1 + x4 + (-1)x2   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   
POL(COND_EVAL_11(x1, x2, x3, x4)) = -1 + x4 + (-1)x2 + x1   

The following pairs are in P>:

COND_EVAL_11(TRUE, x[1], y[1], z[1]) → EVAL_1(x[1], y[1], -@z(x[1], y[1]))

The following pairs are in Pbound:

COND_EVAL_11(TRUE, x[1], y[1], z[1]) → EVAL_1(x[1], y[1], -@z(x[1], y[1]))

The following pairs are in P:

EVAL_1(x[2], y[2], z[2]) → COND_EVAL_11(&&(&&(>@z(y[2], x[2]), >@z(z[2], y[2])), >@z(y[2], 0@z)), x[2], y[2], z[2])
EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])
COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(2): EVAL_1(x[2], y[2], z[2]) → COND_EVAL_11(&&(&&(>@z(y[2], x[2]), >@z(z[2], y[2])), >@z(y[2], 0@z)), x[2], y[2], z[2])
(3): EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])
(0): COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])

(3) -> (0), if ((z[3]* z[0])∧(x[3]* x[0])∧(y[3]* y[0])∧(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)) →* TRUE))


(0) -> (3), if ((y[0]* y[3])∧(z[0]* z[3])∧(+@z(x[0], y[0]) →* x[3]))


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



The set Q consists of the following terms:

eval_1(x0, x1, x2)
Cond_eval_11(TRUE, x0, x1, x2)
Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_0(TRUE, x0, x1, x2)
eval_0(x0, x1, x2)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDependencyGraphProof
            ↳ 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:

(3): EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])
(0): COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])

(3) -> (0), if ((z[3]* z[0])∧(x[3]* x[0])∧(y[3]* y[0])∧(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)) →* TRUE))


(0) -> (3), if ((y[0]* y[3])∧(z[0]* z[3])∧(+@z(x[0], y[0]) →* x[3]))



The set Q consists of the following terms:

eval_1(x0, x1, x2)
Cond_eval_11(TRUE, x0, x1, x2)
Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_0(TRUE, x0, x1, x2)
eval_0(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_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3]) the following chains were created:




For Pair COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), 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(0@z) = 0   
POL(EVAL_1(x1, x2, x3)) = 1 + (2)x3 + (-1)x2 + (-1)x1   
POL(COND_EVAL_1(x1, x2, x3, x4)) = 1 + (2)x4 + (-1)x3 + (-1)x2   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 0   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])

The following pairs are in Pbound:

COND_EVAL_1(TRUE, x[0], y[0], z[0]) → EVAL_1(+@z(x[0], y[0]), y[0], z[0])

The following pairs are in P:

EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])

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
          ↳ IDependencyGraphProof
            ↳ 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:

(3): EVAL_1(x[3], y[3], z[3]) → COND_EVAL_1(&&(&&(>@z(y[3], x[3]), >@z(z[3], y[3])), >@z(y[3], 0@z)), x[3], y[3], z[3])


The set Q consists of the following terms:

eval_1(x0, x1, x2)
Cond_eval_11(TRUE, x0, x1, x2)
Cond_eval_1(TRUE, x0, x1, x2)
Cond_eval_0(TRUE, x0, x1, x2)
eval_0(x0, x1, x2)


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