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:

f(x, y) → Cond_f(>@z(x, y), x, y)
Cond_f(TRUE, x, y) → f(+@z(x, 1@z), +@z(y, 2@z))

The set Q consists of the following terms:

f(x0, x1)
Cond_f(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:

f(x, y) → Cond_f(>@z(x, y), x, y)
Cond_f(TRUE, x, y) → f(+@z(x, 1@z), +@z(y, 2@z))

The integer pair graph contains the following rules and edges:

(0): F(x[0], y[0]) → COND_F(>@z(x[0], y[0]), x[0], y[0])
(1): COND_F(TRUE, x[1], y[1]) → F(+@z(x[1], 1@z), +@z(y[1], 2@z))

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


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



The set Q consists of the following terms:

f(x0, x1)
Cond_f(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): F(x[0], y[0]) → COND_F(>@z(x[0], y[0]), x[0], y[0])
(1): COND_F(TRUE, x[1], y[1]) → F(+@z(x[1], 1@z), +@z(y[1], 2@z))

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


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



The set Q consists of the following terms:

f(x0, x1)
Cond_f(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 F(x, y) → COND_F(>@z(x, y), x, y) the following chains were created:




For Pair COND_F(TRUE, x, y) → F(+@z(x, 1@z), +@z(y, 2@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_F(x1, x2, x3)) = 2 + (-1)x3 + x2   
POL(TRUE) = 0   
POL(2@z) = 2   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(F(x1, x2)) = 2 + (-1)x2 + x1   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = 1   

The following pairs are in P>:

COND_F(TRUE, x[1], y[1]) → F(+@z(x[1], 1@z), +@z(y[1], 2@z))

The following pairs are in Pbound:

COND_F(TRUE, x[1], y[1]) → F(+@z(x[1], 1@z), +@z(y[1], 2@z))

The following pairs are in P:

F(x[0], y[0]) → COND_F(>@z(x[0], y[0]), x[0], y[0])

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

+@z1


↳ 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): F(x[0], y[0]) → COND_F(>@z(x[0], y[0]), x[0], y[0])


The set Q consists of the following terms:

f(x0, x1)
Cond_f(TRUE, x0, x1)


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