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) → Cond_eval3(&&(>@z(y, x), >@z(x, y)), x, y)
eval(x, y) → Cond_eval1(&&(>@z(y, x), >=@z(y, x)), x, y)
eval(x, y) → Cond_eval(&&(>@z(x, y), >=@z(y, x)), x, y)
Cond_eval1(TRUE, x, y) → eval(+@z(x, 1@z), y)
Cond_eval(TRUE, x, y) → eval(+@z(x, 1@z), y)
eval(x, y) → Cond_eval2(>@z(x, y), x, y)
Cond_eval3(TRUE, x, y) → eval(x, +@z(y, 1@z))
Cond_eval2(TRUE, x, y) → eval(x, +@z(y, 1@z))

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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) → Cond_eval3(&&(>@z(y, x), >@z(x, y)), x, y)
eval(x, y) → Cond_eval1(&&(>@z(y, x), >=@z(y, x)), x, y)
eval(x, y) → Cond_eval(&&(>@z(x, y), >=@z(y, x)), x, y)
Cond_eval1(TRUE, x, y) → eval(+@z(x, 1@z), y)
Cond_eval(TRUE, x, y) → eval(+@z(x, 1@z), y)
eval(x, y) → Cond_eval2(>@z(x, y), x, y)
Cond_eval3(TRUE, x, y) → eval(x, +@z(y, 1@z))
Cond_eval2(TRUE, x, y) → eval(x, +@z(y, 1@z))

The integer pair graph contains the following rules and edges:

(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])
(5): COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

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


(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (4), if ((y[0]* y[4])∧(+@z(x[0], 1@z) →* x[4]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


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


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


(1) -> (4), if ((y[1]* y[4])∧(+@z(x[1], 1@z) →* x[4]))


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))


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


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


(5) -> (2), if ((+@z(y[5], 1@z) →* y[2])∧(x[5]* x[2]))


(5) -> (3), if ((+@z(y[5], 1@z) →* y[3])∧(x[5]* x[3]))


(5) -> (4), if ((+@z(y[5], 1@z) →* y[4])∧(x[5]* x[4]))


(5) -> (7), if ((+@z(y[5], 1@z) →* y[7])∧(x[5]* x[7]))


(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(6) -> (4), if ((+@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(6) -> (7), if ((+@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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]) → EVAL(+@z(x[0], 1@z), y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])
(5): COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

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


(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (4), if ((y[0]* y[4])∧(+@z(x[0], 1@z) →* x[4]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


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


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


(1) -> (4), if ((y[1]* y[4])∧(+@z(x[1], 1@z) →* x[4]))


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))


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


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


(5) -> (2), if ((+@z(y[5], 1@z) →* y[2])∧(x[5]* x[2]))


(5) -> (3), if ((+@z(y[5], 1@z) →* y[3])∧(x[5]* x[3]))


(5) -> (4), if ((+@z(y[5], 1@z) →* y[4])∧(x[5]* x[4]))


(5) -> (7), if ((+@z(y[5], 1@z) →* y[7])∧(x[5]* x[7]))


(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(6) -> (4), if ((+@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(6) -> (7), if ((+@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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) → EVAL(+@z(x, 1@z), y) the following chains were created:




For Pair COND_EVAL(TRUE, x, y) → EVAL(+@z(x, 1@z), y) the following chains were created:




For Pair EVAL(x, y) → COND_EVAL2(>@z(x, y), x, y) the following chains were created:




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




For Pair EVAL(x, y) → COND_EVAL3(&&(>@z(y, x), >@z(x, y)), x, y) the following chains were created:




For Pair COND_EVAL3(TRUE, x, y) → EVAL(x, +@z(y, 1@z)) the following chains were created:




For Pair COND_EVAL2(TRUE, x, y) → EVAL(x, +@z(y, 1@z)) the following chains were created:




For Pair EVAL(x, y) → COND_EVAL(&&(>@z(x, y), >=@z(y, x)), x, y) 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)) = 1 + (-1)x3 + (-1)x2 + (2)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (2)x1   
POL(FALSE) = -1   
POL(>@z(x1, x2)) = -1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL3(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2   
POL(COND_EVAL2(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2   
POL(EVAL(x1, x2)) = (-1)x2 + (-1)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))

The following pairs are in P:

COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])
COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))
EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

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

&&(FALSE, FALSE)1FALSE1
&&(TRUE, TRUE)1TRUE1
+@z1
&&(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): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])
(5): COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

(1) -> (4), if ((y[1]* y[4])∧(+@z(x[1], 1@z) →* x[4]))


(6) -> (7), if ((+@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(0) -> (4), if ((y[0]* y[4])∧(+@z(x[0], 1@z) →* x[4]))


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


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


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))


(6) -> (4), if ((+@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


(5) -> (3), if ((+@z(y[5], 1@z) →* y[3])∧(x[5]* x[3]))


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


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


(5) -> (4), if ((+@z(y[5], 1@z) →* y[4])∧(x[5]* x[4]))


(5) -> (7), if ((+@z(y[5], 1@z) →* y[7])∧(x[5]* x[7]))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(5): COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(1) -> (4), if ((y[1]* y[4])∧(+@z(x[1], 1@z) →* x[4]))


(5) -> (3), if ((+@z(y[5], 1@z) →* y[3])∧(x[5]* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


(0) -> (4), if ((y[0]* y[4])∧(+@z(x[0], 1@z) →* x[4]))


(5) -> (4), if ((+@z(y[5], 1@z) →* y[4])∧(x[5]* x[4]))


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


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


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


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


(5) -> (7), if ((+@z(y[5], 1@z) →* y[7])∧(x[5]* x[7]))


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z)) the following chains were created:




For Pair EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3]) the following chains were created:




For Pair COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7]) 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_EVAL3(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2   
POL(COND_EVAL1(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2 + (-1)x1   
POL(EVAL(x1, x2)) = (-1)x2 + (-1)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 0   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])

The following pairs are in Pbound:

COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])

The following pairs are in P:

COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

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

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


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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(5): COND_EVAL3(TRUE, x[5], y[5]) → EVAL(x[5], +@z(y[5], 1@z))
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(5) -> (3), if ((+@z(y[5], 1@z) →* y[3])∧(x[5]* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


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


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


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


(5) -> (7), if ((+@z(y[5], 1@z) →* y[7])∧(x[5]* x[7]))


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


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


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


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


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3]) the following chains were created:




For Pair COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[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)) = -1   
POL(COND_EVAL1(x1, x2, x3)) = -1 + x3 + (-1)x2 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3)) = x3 + (-1)x2 + (2)x1   
POL(EVAL(x1, x2)) = x2 + (-1)x1   
POL(+@z(x1, x2)) = x1 + 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]) → EVAL(+@z(x[1], 1@z), y[1])
EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

The following pairs are in Pbound:

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

The following pairs are in P:

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

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

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


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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

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

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3]) the following chains were created:




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

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])

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

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


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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])


The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

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

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3]) the following chains were created:




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

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])

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

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


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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])


The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ AND
                        ↳ IDP
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:

(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


(0) -> (4), if ((y[0]* y[4])∧(+@z(x[0], 1@z) →* x[4]))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

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

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

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

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3]) the following chains were created:




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

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ AND
                        ↳ IDP
                        ↳ IDP
                          ↳ IDependencyGraphProof
                            ↳ IDP
                              ↳ IDPNonInfProof
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:

(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])


The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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:

z

R is empty.
The integer pair graph contains the following rules and edges:

(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(2): EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(4): EVAL(x[4], y[4]) → COND_EVAL3(&&(>@z(y[4], x[4]), >@z(x[4], y[4])), x[4], y[4])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])

(1) -> (4), if ((y[1]* y[4])∧(+@z(x[1], 1@z) →* x[4]))


(6) -> (7), if ((+@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(0) -> (4), if ((y[0]* y[4])∧(+@z(x[0], 1@z) →* x[4]))


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


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


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))


(6) -> (4), if ((+@z(y[6], 1@z) →* y[4])∧(x[6]* x[4]))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))


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


(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


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


(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ 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:

(2): EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(1): COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1])
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


(6) -> (7), if ((+@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


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


(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(1) -> (7), if ((y[1]* y[7])∧(+@z(x[1], 1@z) →* x[7]))


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


(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(7) -> (1), if ((x[7]* x[1])∧(y[7]* y[1])∧(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])) →* TRUE))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))


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


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3]) the following chains were created:




For Pair COND_EVAL(TRUE, x[1], y[1]) → EVAL(+@z(x[1], 1@z), y[1]) the following chains were created:




For Pair EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7]) the following chains were created:




For Pair COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 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_EVAL2(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2   
POL(COND_EVAL1(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2   
POL(TRUE) = 0   
POL(&&(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3)) = -1 + (-1)x3 + (-1)x2   
POL(EVAL(x1, x2)) = -1 + (-1)x2 + (-1)x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 2   
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]) → EVAL(+@z(x[1], 1@z), y[1])

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])
COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

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
            ↳ AND
              ↳ IDP
              ↳ 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(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(7): EVAL(x[7], y[7]) → COND_EVAL(&&(>@z(x[7], y[7]), >=@z(y[7], x[7])), x[7], y[7])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(0) -> (7), if ((y[0]* y[7])∧(+@z(x[0], 1@z) →* x[7]))


(6) -> (7), if ((+@z(y[6], 1@z) →* y[7])∧(x[6]* x[7]))


(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


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


(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
IDP
                            ↳ IDPtoQDPProof
                            ↳ IDPNonInfProof

I DP problem:
The following domains are used:

z

R is empty.
The integer pair graph contains the following rules and edges:

(2): EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(0): COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

(0) -> (3), if ((y[0]* y[3])∧(+@z(x[0], 1@z) →* x[3]))


(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


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


(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))


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



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


Represented integers and predefined function symbols by Terms

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPtoQDPProof
QDP
                                ↳ UsableRulesProof
                            ↳ IDPNonInfProof

Q DP problem:
The TRS P consists of the following rules:

EVAL(x[2], y[2]) → COND_EVAL2(greater_int(x[2], y[2]), x[2], y[2])
COND_EVAL1(true, x[0], y[0]) → EVAL(plus_int(pos(s(0)), x[0]), y[0])
EVAL(x[3], y[3]) → COND_EVAL1(and(greater_int(y[3], x[3]), greatereq_int(y[3], x[3])), x[3], y[3])
COND_EVAL2(true, x[6], y[6]) → EVAL(x[6], plus_int(pos(s(0)), y[6]))

The TRS R consists of the following rules:

greater_int(pos(0), pos(0)) → false
greater_int(pos(0), neg(0)) → false
greater_int(neg(0), pos(0)) → false
greater_int(neg(0), neg(0)) → false
greater_int(pos(0), pos(s(y))) → false
greater_int(neg(0), pos(s(y))) → false
greater_int(pos(0), neg(s(y))) → true
greater_int(neg(0), neg(s(y))) → true
greater_int(pos(s(x)), pos(0)) → true
greater_int(neg(s(x)), pos(0)) → false
greater_int(pos(s(x)), neg(0)) → true
greater_int(neg(s(x)), neg(0)) → false
greater_int(pos(s(x)), neg(s(y))) → true
greater_int(neg(s(x)), pos(s(y))) → false
greater_int(pos(s(x)), pos(s(y))) → greater_int(pos(x), pos(y))
greater_int(neg(s(x)), neg(s(y))) → greater_int(neg(x), neg(y))
plus_int(pos(x), neg(y)) → minus_nat(x, y)
plus_int(neg(x), pos(y)) → minus_nat(y, x)
plus_int(neg(x), neg(y)) → neg(plus_nat(x, y))
plus_int(pos(x), pos(y)) → pos(plus_nat(x, y))
plus_nat(0, x) → x
plus_nat(s(x), y) → s(plus_nat(x, y))
minus_nat(0, 0) → pos(0)
minus_nat(0, s(y)) → neg(s(y))
minus_nat(s(x), 0) → pos(s(x))
minus_nat(s(x), s(y)) → minus_nat(x, y)
and(false, false) → false
and(false, true) → false
and(true, false) → false
and(true, true) → true
greatereq_int(pos(x), pos(0)) → true
greatereq_int(neg(0), pos(0)) → true
greatereq_int(neg(0), neg(y)) → true
greatereq_int(pos(x), neg(y)) → true
greatereq_int(pos(0), pos(s(y))) → false
greatereq_int(neg(x), pos(s(y))) → false
greatereq_int(neg(s(x)), pos(0)) → false
greatereq_int(neg(s(x)), neg(0)) → false
greatereq_int(pos(s(x)), pos(s(y))) → greatereq_int(pos(x), pos(y))
greatereq_int(neg(s(x)), neg(s(y))) → greatereq_int(neg(x), neg(y))

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(true, x0, x1)
Cond_eval(true, x0, x1)
Cond_eval3(true, x0, x1)
Cond_eval2(true, x0, x1)
greater_int(pos(0), pos(0))
greater_int(pos(0), neg(0))
greater_int(neg(0), pos(0))
greater_int(neg(0), neg(0))
greater_int(pos(0), pos(s(x0)))
greater_int(neg(0), pos(s(x0)))
greater_int(pos(0), neg(s(x0)))
greater_int(neg(0), neg(s(x0)))
greater_int(pos(s(x0)), pos(0))
greater_int(neg(s(x0)), pos(0))
greater_int(pos(s(x0)), neg(0))
greater_int(neg(s(x0)), neg(0))
greater_int(pos(s(x0)), neg(s(x1)))
greater_int(neg(s(x0)), pos(s(x1)))
greater_int(pos(s(x0)), pos(s(x1)))
greater_int(neg(s(x0)), neg(s(x1)))
plus_int(pos(x0), neg(x1))
plus_int(neg(x0), pos(x1))
plus_int(neg(x0), neg(x1))
plus_int(pos(x0), pos(x1))
plus_nat(0, x0)
plus_nat(s(x0), x1)
minus_nat(0, 0)
minus_nat(0, s(x0))
minus_nat(s(x0), 0)
minus_nat(s(x0), s(x1))
and(false, false)
and(false, true)
and(true, false)
and(true, true)
greatereq_int(pos(x0), pos(0))
greatereq_int(neg(0), pos(0))
greatereq_int(neg(0), neg(x0))
greatereq_int(pos(x0), neg(x1))
greatereq_int(pos(0), pos(s(x0)))
greatereq_int(neg(x0), pos(s(x1)))
greatereq_int(neg(s(x0)), pos(0))
greatereq_int(neg(s(x0)), neg(0))
greatereq_int(pos(s(x0)), pos(s(x1)))
greatereq_int(neg(s(x0)), neg(s(x1)))

We have to consider all minimal (P,Q,R)-chains.
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
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPtoQDPProof
                              ↳ QDP
                                ↳ UsableRulesProof
QDP
                                    ↳ QReductionProof
                            ↳ IDPNonInfProof

Q DP problem:
The TRS P consists of the following rules:

EVAL(x[2], y[2]) → COND_EVAL2(greater_int(x[2], y[2]), x[2], y[2])
COND_EVAL1(true, x[0], y[0]) → EVAL(plus_int(pos(s(0)), x[0]), y[0])
EVAL(x[3], y[3]) → COND_EVAL1(and(greater_int(y[3], x[3]), greatereq_int(y[3], x[3])), x[3], y[3])
COND_EVAL2(true, x[6], y[6]) → EVAL(x[6], plus_int(pos(s(0)), y[6]))

The TRS R consists of the following rules:

plus_int(pos(x), neg(y)) → minus_nat(x, y)
plus_int(pos(x), pos(y)) → pos(plus_nat(x, y))
plus_nat(0, x) → x
plus_nat(s(x), y) → s(plus_nat(x, y))
minus_nat(0, 0) → pos(0)
minus_nat(0, s(y)) → neg(s(y))
minus_nat(s(x), 0) → pos(s(x))
minus_nat(s(x), s(y)) → minus_nat(x, y)
greater_int(pos(0), pos(0)) → false
greater_int(pos(0), neg(0)) → false
greater_int(neg(0), pos(0)) → false
greater_int(neg(0), neg(0)) → false
greater_int(pos(0), pos(s(y))) → false
greater_int(neg(0), pos(s(y))) → false
greater_int(pos(0), neg(s(y))) → true
greater_int(neg(0), neg(s(y))) → true
greater_int(pos(s(x)), pos(0)) → true
greater_int(neg(s(x)), pos(0)) → false
greater_int(pos(s(x)), neg(0)) → true
greater_int(neg(s(x)), neg(0)) → false
greater_int(pos(s(x)), neg(s(y))) → true
greater_int(neg(s(x)), pos(s(y))) → false
greater_int(pos(s(x)), pos(s(y))) → greater_int(pos(x), pos(y))
greater_int(neg(s(x)), neg(s(y))) → greater_int(neg(x), neg(y))
greatereq_int(pos(x), pos(0)) → true
greatereq_int(neg(0), pos(0)) → true
greatereq_int(neg(0), neg(y)) → true
greatereq_int(pos(x), neg(y)) → true
greatereq_int(pos(0), pos(s(y))) → false
greatereq_int(neg(x), pos(s(y))) → false
greatereq_int(neg(s(x)), pos(0)) → false
greatereq_int(neg(s(x)), neg(0)) → false
greatereq_int(pos(s(x)), pos(s(y))) → greatereq_int(pos(x), pos(y))
greatereq_int(neg(s(x)), neg(s(y))) → greatereq_int(neg(x), neg(y))
and(false, false) → false
and(false, true) → false
and(true, false) → false
and(true, true) → true

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(true, x0, x1)
Cond_eval(true, x0, x1)
Cond_eval3(true, x0, x1)
Cond_eval2(true, x0, x1)
greater_int(pos(0), pos(0))
greater_int(pos(0), neg(0))
greater_int(neg(0), pos(0))
greater_int(neg(0), neg(0))
greater_int(pos(0), pos(s(x0)))
greater_int(neg(0), pos(s(x0)))
greater_int(pos(0), neg(s(x0)))
greater_int(neg(0), neg(s(x0)))
greater_int(pos(s(x0)), pos(0))
greater_int(neg(s(x0)), pos(0))
greater_int(pos(s(x0)), neg(0))
greater_int(neg(s(x0)), neg(0))
greater_int(pos(s(x0)), neg(s(x1)))
greater_int(neg(s(x0)), pos(s(x1)))
greater_int(pos(s(x0)), pos(s(x1)))
greater_int(neg(s(x0)), neg(s(x1)))
plus_int(pos(x0), neg(x1))
plus_int(neg(x0), pos(x1))
plus_int(neg(x0), neg(x1))
plus_int(pos(x0), pos(x1))
plus_nat(0, x0)
plus_nat(s(x0), x1)
minus_nat(0, 0)
minus_nat(0, s(x0))
minus_nat(s(x0), 0)
minus_nat(s(x0), s(x1))
and(false, false)
and(false, true)
and(true, false)
and(true, true)
greatereq_int(pos(x0), pos(0))
greatereq_int(neg(0), pos(0))
greatereq_int(neg(0), neg(x0))
greatereq_int(pos(x0), neg(x1))
greatereq_int(pos(0), pos(s(x0)))
greatereq_int(neg(x0), pos(s(x1)))
greatereq_int(neg(s(x0)), pos(0))
greatereq_int(neg(s(x0)), neg(0))
greatereq_int(pos(s(x0)), pos(s(x1)))
greatereq_int(neg(s(x0)), neg(s(x1)))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

eval(x0, x1)
Cond_eval1(true, x0, x1)
Cond_eval(true, x0, x1)
Cond_eval3(true, x0, x1)
Cond_eval2(true, x0, x1)



↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPtoQDPProof
                              ↳ QDP
                                ↳ UsableRulesProof
                                  ↳ QDP
                                    ↳ QReductionProof
QDP
                            ↳ IDPNonInfProof

Q DP problem:
The TRS P consists of the following rules:

EVAL(x[2], y[2]) → COND_EVAL2(greater_int(x[2], y[2]), x[2], y[2])
COND_EVAL1(true, x[0], y[0]) → EVAL(plus_int(pos(s(0)), x[0]), y[0])
EVAL(x[3], y[3]) → COND_EVAL1(and(greater_int(y[3], x[3]), greatereq_int(y[3], x[3])), x[3], y[3])
COND_EVAL2(true, x[6], y[6]) → EVAL(x[6], plus_int(pos(s(0)), y[6]))

The TRS R consists of the following rules:

plus_int(pos(x), neg(y)) → minus_nat(x, y)
plus_int(pos(x), pos(y)) → pos(plus_nat(x, y))
plus_nat(0, x) → x
plus_nat(s(x), y) → s(plus_nat(x, y))
minus_nat(0, 0) → pos(0)
minus_nat(0, s(y)) → neg(s(y))
minus_nat(s(x), 0) → pos(s(x))
minus_nat(s(x), s(y)) → minus_nat(x, y)
greater_int(pos(0), pos(0)) → false
greater_int(pos(0), neg(0)) → false
greater_int(neg(0), pos(0)) → false
greater_int(neg(0), neg(0)) → false
greater_int(pos(0), pos(s(y))) → false
greater_int(neg(0), pos(s(y))) → false
greater_int(pos(0), neg(s(y))) → true
greater_int(neg(0), neg(s(y))) → true
greater_int(pos(s(x)), pos(0)) → true
greater_int(neg(s(x)), pos(0)) → false
greater_int(pos(s(x)), neg(0)) → true
greater_int(neg(s(x)), neg(0)) → false
greater_int(pos(s(x)), neg(s(y))) → true
greater_int(neg(s(x)), pos(s(y))) → false
greater_int(pos(s(x)), pos(s(y))) → greater_int(pos(x), pos(y))
greater_int(neg(s(x)), neg(s(y))) → greater_int(neg(x), neg(y))
greatereq_int(pos(x), pos(0)) → true
greatereq_int(neg(0), pos(0)) → true
greatereq_int(neg(0), neg(y)) → true
greatereq_int(pos(x), neg(y)) → true
greatereq_int(pos(0), pos(s(y))) → false
greatereq_int(neg(x), pos(s(y))) → false
greatereq_int(neg(s(x)), pos(0)) → false
greatereq_int(neg(s(x)), neg(0)) → false
greatereq_int(pos(s(x)), pos(s(y))) → greatereq_int(pos(x), pos(y))
greatereq_int(neg(s(x)), neg(s(y))) → greatereq_int(neg(x), neg(y))
and(false, false) → false
and(false, true) → false
and(true, false) → false
and(true, true) → true

The set Q consists of the following terms:

greater_int(pos(0), pos(0))
greater_int(pos(0), neg(0))
greater_int(neg(0), pos(0))
greater_int(neg(0), neg(0))
greater_int(pos(0), pos(s(x0)))
greater_int(neg(0), pos(s(x0)))
greater_int(pos(0), neg(s(x0)))
greater_int(neg(0), neg(s(x0)))
greater_int(pos(s(x0)), pos(0))
greater_int(neg(s(x0)), pos(0))
greater_int(pos(s(x0)), neg(0))
greater_int(neg(s(x0)), neg(0))
greater_int(pos(s(x0)), neg(s(x1)))
greater_int(neg(s(x0)), pos(s(x1)))
greater_int(pos(s(x0)), pos(s(x1)))
greater_int(neg(s(x0)), neg(s(x1)))
plus_int(pos(x0), neg(x1))
plus_int(neg(x0), pos(x1))
plus_int(neg(x0), neg(x1))
plus_int(pos(x0), pos(x1))
plus_nat(0, x0)
plus_nat(s(x0), x1)
minus_nat(0, 0)
minus_nat(0, s(x0))
minus_nat(s(x0), 0)
minus_nat(s(x0), s(x1))
and(false, false)
and(false, true)
and(true, false)
and(true, true)
greatereq_int(pos(x0), pos(0))
greatereq_int(neg(0), pos(0))
greatereq_int(neg(0), neg(x0))
greatereq_int(pos(x0), neg(x1))
greatereq_int(pos(0), pos(s(x0)))
greatereq_int(neg(x0), pos(s(x1)))
greatereq_int(neg(s(x0)), pos(0))
greatereq_int(neg(s(x0)), neg(0))
greatereq_int(pos(s(x0)), pos(s(x1)))
greatereq_int(neg(s(x0)), neg(s(x1)))

We have to consider all minimal (P,Q,R)-chains.
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[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2]) the following chains were created:




For Pair COND_EVAL1(TRUE, x[0], y[0]) → EVAL(+@z(x[0], 1@z), y[0]) the following chains were created:




For Pair EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3]) the following chains were created:




For Pair COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 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_EVAL2(x1, x2, x3)) = -1 + max{(-1)x2 + x3, x2 + (-1)x3}   
POL(COND_EVAL1(x1, x2, x3)) = -1 + max{(-1)x2 + x3, x2 + (-1)x3}   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(EVAL(x1, x2)) = -1 + max{(-1)x1 + x2, x1 + (-1)x2}   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
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]) → EVAL(+@z(x[0], 1@z), y[0])

The following pairs are in Pbound:

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

The following pairs are in P:

EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPtoQDPProof
                            ↳ 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(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(3): EVAL(x[3], y[3]) → COND_EVAL1(&&(>@z(y[3], x[3]), >=@z(y[3], x[3])), x[3], y[3])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

(6) -> (3), if ((+@z(y[6], 1@z) →* y[3])∧(x[6]* x[3]))


(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPtoQDPProof
                            ↳ 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:

(2): EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])
(6): COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

(6) -> (2), if ((+@z(y[6], 1@z) →* y[2])∧(x[6]* x[2]))


(2) -> (6), if ((x[2]* x[6])∧(y[2]* y[6])∧(>@z(x[2], y[2]) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2]) the following chains were created:




For Pair COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 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(COND_EVAL2(x1, x2, x3)) = -1 + (-1)x3 + x2   
POL(TRUE) = 2   
POL(EVAL(x1, x2)) = -1 + (-1)x2 + x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

The following pairs are in Pbound:

COND_EVAL2(TRUE, x[6], y[6]) → EVAL(x[6], +@z(y[6], 1@z))

The following pairs are in P:

EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])

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

+@z1


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ IDP
                ↳ IDependencyGraphProof
                  ↳ IDP
                    ↳ IDPNonInfProof
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPtoQDPProof
                            ↳ 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:

(2): EVAL(x[2], y[2]) → COND_EVAL2(>@z(x[2], y[2]), x[2], y[2])


The set Q consists of the following terms:

eval(x0, x1)
Cond_eval1(TRUE, x0, x1)
Cond_eval(TRUE, x0, x1)
Cond_eval3(TRUE, x0, x1)
Cond_eval2(TRUE, x0, x1)


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