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(i, j) → Cond_eval(&&(&&(>=@z(-@z(i, j), 1@z), >=@z(nat, 0@z)), >@z(pos, 0@z)), i, j, nat, pos)
Cond_eval(TRUE, i, j, nat, pos) → eval(-@z(i, nat), +@z(j, pos))

The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(TRUE, x0, x1, x2, x3)


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(i, j) → Cond_eval(&&(&&(>=@z(-@z(i, j), 1@z), >=@z(nat, 0@z)), >@z(pos, 0@z)), i, j, nat, pos)
Cond_eval(TRUE, i, j, nat, pos) → eval(-@z(i, nat), +@z(j, pos))

The integer pair graph contains the following rules and edges:

(0): EVAL(i[0], j[0]) → COND_EVAL(&&(&&(>=@z(-@z(i[0], j[0]), 1@z), >=@z(nat[0], 0@z)), >@z(pos[0], 0@z)), i[0], j[0], nat[0], pos[0])
(1): COND_EVAL(TRUE, i[1], j[1], nat[1], pos[1]) → EVAL(-@z(i[1], nat[1]), +@z(j[1], pos[1]))

(0) -> (1), if ((nat[0]* nat[1])∧(pos[0]* pos[1])∧(i[0]* i[1])∧(j[0]* j[1])∧(&&(&&(>=@z(-@z(i[0], j[0]), 1@z), >=@z(nat[0], 0@z)), >@z(pos[0], 0@z)) →* TRUE))


(1) -> (0), if ((+@z(j[1], pos[1]) →* j[0])∧(-@z(i[1], nat[1]) →* i[0]))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(TRUE, x0, x1, x2, x3)


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): EVAL(i[0], j[0]) → COND_EVAL(&&(&&(>=@z(-@z(i[0], j[0]), 1@z), >=@z(nat[0], 0@z)), >@z(pos[0], 0@z)), i[0], j[0], nat[0], pos[0])
(1): COND_EVAL(TRUE, i[1], j[1], nat[1], pos[1]) → EVAL(-@z(i[1], nat[1]), +@z(j[1], pos[1]))

(0) -> (1), if ((nat[0]* nat[1])∧(pos[0]* pos[1])∧(i[0]* i[1])∧(j[0]* j[1])∧(&&(&&(>=@z(-@z(i[0], j[0]), 1@z), >=@z(nat[0], 0@z)), >@z(pos[0], 0@z)) →* TRUE))


(1) -> (0), if ((+@z(j[1], pos[1]) →* j[0])∧(-@z(i[1], nat[1]) →* i[0]))



The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(TRUE, x0, x1, x2, x3)


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(i, j) → COND_EVAL(&&(&&(>=@z(-@z(i, j), 1@z), >=@z(nat, 0@z)), >@z(pos, 0@z)), i, j, nat, pos) the following chains were created:




For Pair COND_EVAL(TRUE, i, j, nat, pos) → EVAL(-@z(i, nat), +@z(j, pos)) the following chains were created:




To summarize, we get the following constraints P for the following pairs.



The constraints for P> respective Pbound are constructed from P where we just replace every occurence of "t ≥ s" in P by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation over integers[POLO]:

POL(-@z(x1, x2)) = x1 + (-1)x2   
POL(>=@z(x1, x2)) = -1   
POL(0@z) = 0   
POL(TRUE) = 0   
POL(&&(x1, x2)) = 0   
POL(EVAL(x1, x2)) = -1 + (-1)x2 + x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(COND_EVAL(x1, x2, x3, x4, x5)) = -1 + (-1)x3 + x2 + (-1)x1   
POL(FALSE) = 1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_EVAL(TRUE, i[1], j[1], nat[1], pos[1]) → EVAL(-@z(i[1], nat[1]), +@z(j[1], pos[1]))

The following pairs are in Pbound:

COND_EVAL(TRUE, i[1], j[1], nat[1], pos[1]) → EVAL(-@z(i[1], nat[1]), +@z(j[1], pos[1]))

The following pairs are in P:

EVAL(i[0], j[0]) → COND_EVAL(&&(&&(>=@z(-@z(i[0], j[0]), 1@z), >=@z(nat[0], 0@z)), >@z(pos[0], 0@z)), i[0], j[0], nat[0], pos[0])

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

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


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

I DP problem:
The following domains are used:

z

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

(0): EVAL(i[0], j[0]) → COND_EVAL(&&(&&(>=@z(-@z(i[0], j[0]), 1@z), >=@z(nat[0], 0@z)), >@z(pos[0], 0@z)), i[0], j[0], nat[0], pos[0])


The set Q consists of the following terms:

eval(x0, x1)
Cond_eval(TRUE, x0, x1, x2, x3)


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