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

The set Q consists of the following terms:

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

The integer pair graph contains the following rules and edges:

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

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



The set Q consists of the following terms:

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

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



The set Q consists of the following terms:

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(TRUE, x, y) → F(&&(>@z(x, y), >@z(y, 2@z)), +@z(1@z, x), *@z(2@z, 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(*@z(x1, x2)) = x1·x2   
POL(TRUE) = -1   
POL(&&(x1, x2)) = 1   
POL(2@z) = 2   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = 1   
POL(F(x1, x2, x3)) = 1 + (-1)x3 + x2   
POL(1@z) = 1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

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

The following pairs are in Pbound:

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

The following pairs are in P:
none

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

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


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

I DP problem:
The following domains are used:none

R is empty.
The integer pair graph is empty.
The set Q consists of the following terms:

f(TRUE, x0, x1)


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