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(an, bn, cn, i, j) → Cond_eval2(&&(<@z(j, bn), <@z(i, an)), an, bn, cn, i, j)
eval(an, bn, cn, i, j) → Cond_eval1(&&(>=@z(j, bn), <@z(i, an)), an, bn, cn, i, j)
Cond_eval(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), +@z(i, 1@z), j)
eval(an, bn, cn, i, j) → Cond_eval(&&(<@z(j, bn), <@z(i, an)), an, bn, cn, i, j)
Cond_eval2(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), i, +@z(j, 1@z))
eval(an, bn, cn, i, j) → Cond_eval3(&&(<@z(j, bn), >=@z(i, an)), an, bn, cn, i, j)
Cond_eval3(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), i, +@z(j, 1@z))
Cond_eval1(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), +@z(i, 1@z), j)

The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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(an, bn, cn, i, j) → Cond_eval2(&&(<@z(j, bn), <@z(i, an)), an, bn, cn, i, j)
eval(an, bn, cn, i, j) → Cond_eval1(&&(>=@z(j, bn), <@z(i, an)), an, bn, cn, i, j)
Cond_eval(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), +@z(i, 1@z), j)
eval(an, bn, cn, i, j) → Cond_eval(&&(<@z(j, bn), <@z(i, an)), an, bn, cn, i, j)
Cond_eval2(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), i, +@z(j, 1@z))
eval(an, bn, cn, i, j) → Cond_eval3(&&(<@z(j, bn), >=@z(i, an)), an, bn, cn, i, j)
Cond_eval3(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), i, +@z(j, 1@z))
Cond_eval1(TRUE, an, bn, cn, i, j) → eval(an, bn, +@z(cn, 1@z), +@z(i, 1@z), j)

The integer pair graph contains the following rules and edges:

(0): COND_EVAL(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])
(1): EVAL(an[1], bn[1], cn[1], i[1], j[1]) → COND_EVAL2(&&(<@z(j[1], bn[1]), <@z(i[1], an[1])), an[1], bn[1], cn[1], i[1], j[1])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(5): COND_EVAL2(TRUE, an[5], bn[5], cn[5], i[5], j[5]) → EVAL(an[5], bn[5], +@z(cn[5], 1@z), i[5], +@z(j[5], 1@z))
(6): EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6])
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

(0) -> (1), if ((+@z(i[0], 1@z) →* i[1])∧(j[0]* j[1])∧(bn[0]* bn[1])∧(+@z(cn[0], 1@z) →* cn[1])∧(an[0]* an[1]))


(0) -> (2), if ((+@z(i[0], 1@z) →* i[2])∧(j[0]* j[2])∧(bn[0]* bn[2])∧(+@z(cn[0], 1@z) →* cn[2])∧(an[0]* an[2]))


(0) -> (6), if ((+@z(i[0], 1@z) →* i[6])∧(j[0]* j[6])∧(bn[0]* bn[6])∧(+@z(cn[0], 1@z) →* cn[6])∧(an[0]* an[6]))


(0) -> (7), if ((+@z(i[0], 1@z) →* i[7])∧(j[0]* j[7])∧(bn[0]* bn[7])∧(+@z(cn[0], 1@z) →* cn[7])∧(an[0]* an[7]))


(1) -> (5), if ((cn[1]* cn[5])∧(i[1]* i[5])∧(an[1]* an[5])∧(bn[1]* bn[5])∧(j[1]* j[5])∧(&&(<@z(j[1], bn[1]), <@z(i[1], an[1])) →* TRUE))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(3) -> (1), if ((i[3]* i[1])∧(+@z(j[3], 1@z) →* j[1])∧(bn[3]* bn[1])∧(+@z(cn[3], 1@z) →* cn[1])∧(an[3]* an[1]))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))


(3) -> (6), if ((i[3]* i[6])∧(+@z(j[3], 1@z) →* j[6])∧(bn[3]* bn[6])∧(+@z(cn[3], 1@z) →* cn[6])∧(an[3]* an[6]))


(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(4) -> (1), if ((+@z(i[4], 1@z) →* i[1])∧(j[4]* j[1])∧(bn[4]* bn[1])∧(+@z(cn[4], 1@z) →* cn[1])∧(an[4]* an[1]))


(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(4) -> (6), if ((+@z(i[4], 1@z) →* i[6])∧(j[4]* j[6])∧(bn[4]* bn[6])∧(+@z(cn[4], 1@z) →* cn[6])∧(an[4]* an[6]))


(4) -> (7), if ((+@z(i[4], 1@z) →* i[7])∧(j[4]* j[7])∧(bn[4]* bn[7])∧(+@z(cn[4], 1@z) →* cn[7])∧(an[4]* an[7]))


(5) -> (1), if ((i[5]* i[1])∧(+@z(j[5], 1@z) →* j[1])∧(bn[5]* bn[1])∧(+@z(cn[5], 1@z) →* cn[1])∧(an[5]* an[1]))


(5) -> (2), if ((i[5]* i[2])∧(+@z(j[5], 1@z) →* j[2])∧(bn[5]* bn[2])∧(+@z(cn[5], 1@z) →* cn[2])∧(an[5]* an[2]))


(5) -> (6), if ((i[5]* i[6])∧(+@z(j[5], 1@z) →* j[6])∧(bn[5]* bn[6])∧(+@z(cn[5], 1@z) →* cn[6])∧(an[5]* an[6]))


(5) -> (7), if ((i[5]* i[7])∧(+@z(j[5], 1@z) →* j[7])∧(bn[5]* bn[7])∧(+@z(cn[5], 1@z) →* cn[7])∧(an[5]* an[7]))


(6) -> (0), if ((cn[6]* cn[0])∧(i[6]* i[0])∧(an[6]* an[0])∧(bn[6]* bn[0])∧(j[6]* j[0])∧(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])) →* TRUE))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])
(1): EVAL(an[1], bn[1], cn[1], i[1], j[1]) → COND_EVAL2(&&(<@z(j[1], bn[1]), <@z(i[1], an[1])), an[1], bn[1], cn[1], i[1], j[1])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(5): COND_EVAL2(TRUE, an[5], bn[5], cn[5], i[5], j[5]) → EVAL(an[5], bn[5], +@z(cn[5], 1@z), i[5], +@z(j[5], 1@z))
(6): EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6])
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

(0) -> (1), if ((+@z(i[0], 1@z) →* i[1])∧(j[0]* j[1])∧(bn[0]* bn[1])∧(+@z(cn[0], 1@z) →* cn[1])∧(an[0]* an[1]))


(0) -> (2), if ((+@z(i[0], 1@z) →* i[2])∧(j[0]* j[2])∧(bn[0]* bn[2])∧(+@z(cn[0], 1@z) →* cn[2])∧(an[0]* an[2]))


(0) -> (6), if ((+@z(i[0], 1@z) →* i[6])∧(j[0]* j[6])∧(bn[0]* bn[6])∧(+@z(cn[0], 1@z) →* cn[6])∧(an[0]* an[6]))


(0) -> (7), if ((+@z(i[0], 1@z) →* i[7])∧(j[0]* j[7])∧(bn[0]* bn[7])∧(+@z(cn[0], 1@z) →* cn[7])∧(an[0]* an[7]))


(1) -> (5), if ((cn[1]* cn[5])∧(i[1]* i[5])∧(an[1]* an[5])∧(bn[1]* bn[5])∧(j[1]* j[5])∧(&&(<@z(j[1], bn[1]), <@z(i[1], an[1])) →* TRUE))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(3) -> (1), if ((i[3]* i[1])∧(+@z(j[3], 1@z) →* j[1])∧(bn[3]* bn[1])∧(+@z(cn[3], 1@z) →* cn[1])∧(an[3]* an[1]))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))


(3) -> (6), if ((i[3]* i[6])∧(+@z(j[3], 1@z) →* j[6])∧(bn[3]* bn[6])∧(+@z(cn[3], 1@z) →* cn[6])∧(an[3]* an[6]))


(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(4) -> (1), if ((+@z(i[4], 1@z) →* i[1])∧(j[4]* j[1])∧(bn[4]* bn[1])∧(+@z(cn[4], 1@z) →* cn[1])∧(an[4]* an[1]))


(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(4) -> (6), if ((+@z(i[4], 1@z) →* i[6])∧(j[4]* j[6])∧(bn[4]* bn[6])∧(+@z(cn[4], 1@z) →* cn[6])∧(an[4]* an[6]))


(4) -> (7), if ((+@z(i[4], 1@z) →* i[7])∧(j[4]* j[7])∧(bn[4]* bn[7])∧(+@z(cn[4], 1@z) →* cn[7])∧(an[4]* an[7]))


(5) -> (1), if ((i[5]* i[1])∧(+@z(j[5], 1@z) →* j[1])∧(bn[5]* bn[1])∧(+@z(cn[5], 1@z) →* cn[1])∧(an[5]* an[1]))


(5) -> (2), if ((i[5]* i[2])∧(+@z(j[5], 1@z) →* j[2])∧(bn[5]* bn[2])∧(+@z(cn[5], 1@z) →* cn[2])∧(an[5]* an[2]))


(5) -> (6), if ((i[5]* i[6])∧(+@z(j[5], 1@z) →* j[6])∧(bn[5]* bn[6])∧(+@z(cn[5], 1@z) →* cn[6])∧(an[5]* an[6]))


(5) -> (7), if ((i[5]* i[7])∧(+@z(j[5], 1@z) →* j[7])∧(bn[5]* bn[7])∧(+@z(cn[5], 1@z) →* cn[7])∧(an[5]* an[7]))


(6) -> (0), if ((cn[6]* cn[0])∧(i[6]* i[0])∧(an[6]* an[0])∧(bn[6]* bn[0])∧(j[6]* j[0])∧(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])) →* TRUE))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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, an, bn, cn, i, j) → EVAL(an, bn, +@z(cn, 1@z), +@z(i, 1@z), j) the following chains were created:




For Pair EVAL(an, bn, cn, i, j) → COND_EVAL2(&&(<@z(j, bn), <@z(i, an)), an, bn, cn, i, j) the following chains were created:




For Pair EVAL(an, bn, cn, i, j) → COND_EVAL1(&&(>=@z(j, bn), <@z(i, an)), an, bn, cn, i, j) the following chains were created:




For Pair COND_EVAL3(TRUE, an, bn, cn, i, j) → EVAL(an, bn, +@z(cn, 1@z), i, +@z(j, 1@z)) the following chains were created:




For Pair COND_EVAL1(TRUE, an, bn, cn, i, j) → EVAL(an, bn, +@z(cn, 1@z), +@z(i, 1@z), j) the following chains were created:




For Pair COND_EVAL2(TRUE, an, bn, cn, i, j) → EVAL(an, bn, +@z(cn, 1@z), i, +@z(j, 1@z)) the following chains were created:




For Pair EVAL(an, bn, cn, i, j) → COND_EVAL(&&(<@z(j, bn), <@z(i, an)), an, bn, cn, i, j) the following chains were created:




For Pair EVAL(an, bn, cn, i, j) → COND_EVAL3(&&(<@z(j, bn), >=@z(i, an)), an, bn, cn, i, j) 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, x4, x5, x6)) = -1 + (-1)x6 + (-1)x5 + x3 + x2   
POL(COND_EVAL3(x1, x2, x3, x4, x5, x6)) = 1 + (-1)x6 + (-1)x5 + x3 + x2 + (2)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(FALSE) = -1   
POL(<@z(x1, x2)) = -1   
POL(EVAL(x1, x2, x3, x4, x5)) = -1 + (-1)x5 + (-1)x4 + x2 + x1   
POL(COND_EVAL1(x1, x2, x3, x4, x5, x6)) = -1 + (-1)x6 + (-1)x5 + x3 + x2 + x1   
POL(>=@z(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND_EVAL(x1, x2, x3, x4, x5, x6)) = -1 + (-1)x6 + (-1)x5 + x3 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL2(TRUE, an[5], bn[5], cn[5], i[5], j[5]) → EVAL(an[5], bn[5], +@z(cn[5], 1@z), i[5], +@z(j[5], 1@z))

The following pairs are in Pbound:

COND_EVAL2(TRUE, an[5], bn[5], cn[5], i[5], j[5]) → EVAL(an[5], bn[5], +@z(cn[5], 1@z), i[5], +@z(j[5], 1@z))

The following pairs are in P:

COND_EVAL(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])
EVAL(an[1], bn[1], cn[1], i[1], j[1]) → COND_EVAL2(&&(<@z(j[1], bn[1]), <@z(i[1], an[1])), an[1], bn[1], cn[1], i[1], j[1])
EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6])
EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[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
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(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])
(1): EVAL(an[1], bn[1], cn[1], i[1], j[1]) → COND_EVAL2(&&(<@z(j[1], bn[1]), <@z(i[1], an[1])), an[1], bn[1], cn[1], i[1], j[1])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(6): EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6])
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

(0) -> (2), if ((+@z(i[0], 1@z) →* i[2])∧(j[0]* j[2])∧(bn[0]* bn[2])∧(+@z(cn[0], 1@z) →* cn[2])∧(an[0]* an[2]))


(4) -> (6), if ((+@z(i[4], 1@z) →* i[6])∧(j[4]* j[6])∧(bn[4]* bn[6])∧(+@z(cn[4], 1@z) →* cn[6])∧(an[4]* an[6]))


(6) -> (0), if ((cn[6]* cn[0])∧(i[6]* i[0])∧(an[6]* an[0])∧(bn[6]* bn[0])∧(j[6]* j[0])∧(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])) →* TRUE))


(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(0) -> (6), if ((+@z(i[0], 1@z) →* i[6])∧(j[0]* j[6])∧(bn[0]* bn[6])∧(+@z(cn[0], 1@z) →* cn[6])∧(an[0]* an[6]))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))


(0) -> (7), if ((+@z(i[0], 1@z) →* i[7])∧(j[0]* j[7])∧(bn[0]* bn[7])∧(+@z(cn[0], 1@z) →* cn[7])∧(an[0]* an[7]))


(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(4) -> (7), if ((+@z(i[4], 1@z) →* i[7])∧(j[4]* j[7])∧(bn[4]* bn[7])∧(+@z(cn[4], 1@z) →* cn[7])∧(an[4]* an[7]))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))


(0) -> (1), if ((+@z(i[0], 1@z) →* i[1])∧(j[0]* j[1])∧(bn[0]* bn[1])∧(+@z(cn[0], 1@z) →* cn[1])∧(an[0]* an[1]))


(3) -> (6), if ((i[3]* i[6])∧(+@z(j[3], 1@z) →* j[6])∧(bn[3]* bn[6])∧(+@z(cn[3], 1@z) →* cn[6])∧(an[3]* an[6]))


(4) -> (1), if ((+@z(i[4], 1@z) →* i[1])∧(j[4]* j[1])∧(bn[4]* bn[1])∧(+@z(cn[4], 1@z) →* cn[1])∧(an[4]* an[1]))


(3) -> (1), if ((i[3]* i[1])∧(+@z(j[3], 1@z) →* j[1])∧(bn[3]* bn[1])∧(+@z(cn[3], 1@z) →* cn[1])∧(an[3]* an[1]))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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:

(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])
(0): COND_EVAL(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])
(6): EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6])

(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(0) -> (2), if ((+@z(i[0], 1@z) →* i[2])∧(j[0]* j[2])∧(bn[0]* bn[2])∧(+@z(cn[0], 1@z) →* cn[2])∧(an[0]* an[2]))


(4) -> (6), if ((+@z(i[4], 1@z) →* i[6])∧(j[4]* j[6])∧(bn[4]* bn[6])∧(+@z(cn[4], 1@z) →* cn[6])∧(an[4]* an[6]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(6) -> (0), if ((cn[6]* cn[0])∧(i[6]* i[0])∧(an[6]* an[0])∧(bn[6]* bn[0])∧(j[6]* j[0])∧(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])) →* TRUE))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))


(4) -> (7), if ((+@z(i[4], 1@z) →* i[7])∧(j[4]* j[7])∧(bn[4]* bn[7])∧(+@z(cn[4], 1@z) →* cn[7])∧(an[4]* an[7]))


(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(3) -> (6), if ((i[3]* i[6])∧(+@z(j[3], 1@z) →* j[6])∧(bn[3]* bn[6])∧(+@z(cn[3], 1@z) →* cn[6])∧(an[3]* an[6]))


(0) -> (6), if ((+@z(i[0], 1@z) →* i[6])∧(j[0]* j[6])∧(bn[0]* bn[6])∧(+@z(cn[0], 1@z) →* cn[6])∧(an[0]* an[6]))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))


(0) -> (7), if ((+@z(i[0], 1@z) →* i[7])∧(j[0]* j[7])∧(bn[0]* bn[7])∧(+@z(cn[0], 1@z) →* cn[7])∧(an[0]* an[7]))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4]) the following chains were created:




For Pair EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2]) the following chains were created:




For Pair COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z)) the following chains were created:




For Pair EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7]) the following chains were created:




For Pair COND_EVAL(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0]) the following chains were created:




For Pair EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6]) 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_EVAL3(x1, x2, x3, x4, x5, x6)) = -1 + (-1)x6 + (-1)x5 + x3 + x2   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4, x5, x6)) = -1 + (-1)x6 + (-1)x5 + x3 + x2 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(<@z(x1, x2)) = -1   
POL(COND_EVAL(x1, x2, x3, x4, x5, x6)) = 1 + (-1)x6 + (-1)x5 + x3 + x2 + (2)x1   
POL(EVAL(x1, x2, x3, x4, x5)) = (-1)x5 + (-1)x4 + x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6])

The following pairs are in Pbound:

COND_EVAL(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])

The following pairs are in P:

EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])
COND_EVAL(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])

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
            ↳ IDP
              ↳ IDependencyGraphProof
                ↳ 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:

(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])
(6): EVAL(an[6], bn[6], cn[6], i[6], j[6]) → COND_EVAL(&&(<@z(j[6], bn[6]), <@z(i[6], an[6])), an[6], bn[6], cn[6], i[6], j[6])

(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(4) -> (6), if ((+@z(i[4], 1@z) →* i[6])∧(j[4]* j[6])∧(bn[4]* bn[6])∧(+@z(cn[4], 1@z) →* cn[6])∧(an[4]* an[6]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))


(4) -> (7), if ((+@z(i[4], 1@z) →* i[7])∧(j[4]* j[7])∧(bn[4]* bn[7])∧(+@z(cn[4], 1@z) →* cn[7])∧(an[4]* an[7]))


(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(3) -> (6), if ((i[3]* i[6])∧(+@z(j[3], 1@z) →* j[6])∧(bn[3]* bn[6])∧(+@z(cn[3], 1@z) →* cn[6])∧(an[3]* an[6]))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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
                    ↳ 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:

(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))


(4) -> (7), if ((+@z(i[4], 1@z) →* i[7])∧(j[4]* j[7])∧(bn[4]* bn[7])∧(+@z(cn[4], 1@z) →* cn[7])∧(an[4]* an[7]))


(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4]) the following chains were created:




For Pair EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2]) the following chains were created:




For Pair COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z)) the following chains were created:




For Pair EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[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(COND_EVAL3(x1, x2, x3, x4, x5, x6)) = -1 + (-1)x6 + x3 + x1   
POL(>=@z(x1, x2)) = -1   
POL(COND_EVAL1(x1, x2, x3, x4, x5, x6)) = -1 + (-1)x6 + x3   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(<@z(x1, x2)) = -1   
POL(EVAL(x1, x2, x3, x4, x5)) = -1 + (-1)x5 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

The following pairs are in Pbound:

COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))

The following pairs are in P:

COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))

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
          ↳ IDPNonInfProof
            ↳ IDP
              ↳ IDependencyGraphProof
                ↳ 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:

(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))

(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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
                    ↳ 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:

(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])

(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4]) the following chains were created:




For Pair EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2]) 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, x5, x6)) = -1 + x6 + (-1)x5 + (-1)x3 + x2 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 2   
POL(<@z(x1, x2)) = -1   
POL(EVAL(x1, x2, x3, x4, x5)) = x5 + (-1)x4 + (-1)x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])

The following pairs are in Pbound:

COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])

The following pairs are in P:

EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])

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
                ↳ IDP
                  ↳ IDPNonInfProof
                    ↳ AND
                      ↳ IDP
                        ↳ IDependencyGraphProof
                          ↳ IDP
                            ↳ IDPNonInfProof
                              ↳ AND
                                ↳ IDP
                                  ↳ IDependencyGraphProof
                                    ↳ 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:

(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])


The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ IDP
              ↳ IDependencyGraphProof
                ↳ 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): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])
(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))


(4) -> (7), if ((+@z(i[4], 1@z) →* i[7])∧(j[4]* j[7])∧(bn[4]* bn[7])∧(+@z(cn[4], 1@z) →* cn[7])∧(an[4]* an[7]))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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
                    ↳ 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:

(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(4): COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])

(4) -> (2), if ((+@z(i[4], 1@z) →* i[2])∧(j[4]* j[2])∧(bn[4]* bn[2])∧(+@z(cn[4], 1@z) →* cn[2])∧(an[4]* an[2]))


(2) -> (4), if ((cn[2]* cn[4])∧(i[2]* i[4])∧(an[2]* an[4])∧(bn[2]* bn[4])∧(j[2]* j[4])∧(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2]) the following chains were created:




For Pair COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4]) 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, x5, x6)) = x6 + (-1)x5 + (-1)x3 + x2 + (-1)x1   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 1   
POL(<@z(x1, x2)) = -1   
POL(EVAL(x1, x2, x3, x4, x5)) = -1 + x5 + (-1)x4 + (-1)x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])

The following pairs are in Pbound:

COND_EVAL1(TRUE, an[4], bn[4], cn[4], i[4], j[4]) → EVAL(an[4], bn[4], +@z(cn[4], 1@z), +@z(i[4], 1@z), j[4])

The following pairs are in P:

EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])

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

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


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

(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])


The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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

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

(2): EVAL(an[2], bn[2], cn[2], i[2], j[2]) → COND_EVAL1(&&(>=@z(j[2], bn[2]), <@z(i[2], an[2])), an[2], bn[2], cn[2], i[2], j[2])
(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])
(0): COND_EVAL(TRUE, an[0], bn[0], cn[0], i[0], j[0]) → EVAL(an[0], bn[0], +@z(cn[0], 1@z), +@z(i[0], 1@z), j[0])

(0) -> (2), if ((+@z(i[0], 1@z) →* i[2])∧(j[0]* j[2])∧(bn[0]* bn[2])∧(+@z(cn[0], 1@z) →* cn[2])∧(an[0]* an[2]))


(3) -> (2), if ((i[3]* i[2])∧(+@z(j[3], 1@z) →* j[2])∧(bn[3]* bn[2])∧(+@z(cn[3], 1@z) →* cn[2])∧(an[3]* an[2]))


(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))


(0) -> (7), if ((+@z(i[0], 1@z) →* i[7])∧(j[0]* j[7])∧(bn[0]* bn[7])∧(+@z(cn[0], 1@z) →* cn[7])∧(an[0]* an[7]))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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

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

(3): COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))
(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

(3) -> (7), if ((i[3]* i[7])∧(+@z(j[3], 1@z) →* j[7])∧(bn[3]* bn[7])∧(+@z(cn[3], 1@z) →* cn[7])∧(an[3]* an[7]))


(7) -> (3), if ((cn[7]* cn[3])∧(i[7]* i[3])∧(an[7]* an[3])∧(bn[7]* bn[3])∧(j[7]* j[3])∧(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])) →* TRUE))



The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z)) the following chains were created:




For Pair EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[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, x4, x5, x6)) = -1 + (-1)x6 + x5 + x3 + (-1)x2 + (-1)x1   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 2   
POL(<@z(x1, x2)) = -1   
POL(EVAL(x1, x2, x3, x4, x5)) = (-1)x5 + x4 + x2 + (-1)x1   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))

The following pairs are in Pbound:

COND_EVAL3(TRUE, an[3], bn[3], cn[3], i[3], j[3]) → EVAL(an[3], bn[3], +@z(cn[3], 1@z), i[3], +@z(j[3], 1@z))

The following pairs are in P:

EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])

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

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


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

(7): EVAL(an[7], bn[7], cn[7], i[7], j[7]) → COND_EVAL3(&&(<@z(j[7], bn[7]), >=@z(i[7], an[7])), an[7], bn[7], cn[7], i[7], j[7])


The set Q consists of the following terms:

eval(x0, x1, x2, x3, x4)
Cond_eval(TRUE, x0, x1, x2, x3, x4)
Cond_eval2(TRUE, x0, x1, x2, x3, x4)
Cond_eval3(TRUE, x0, x1, x2, x3, x4)
Cond_eval1(TRUE, x0, x1, x2, x3, x4)


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