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:

cond(TRUE, x, y) → +@z(1@z, minus(x, +@z(y, 1@z)))
cond(FALSE, x, y) → 0@z
minus(x, y) → cond(>=@z(x, +@z(y, 1@z)), x, y)

The set Q consists of the following terms:

cond(TRUE, x0, x1)
cond(FALSE, x0, x1)
minus(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:

cond(TRUE, x, y) → +@z(1@z, minus(x, +@z(y, 1@z)))
cond(FALSE, x, y) → 0@z
minus(x, y) → cond(>=@z(x, +@z(y, 1@z)), x, y)

The integer pair graph contains the following rules and edges:

(0): MINUS(x[0], y[0]) → COND(>=@z(x[0], +@z(y[0], 1@z)), x[0], y[0])
(1): COND(TRUE, x[1], y[1]) → MINUS(x[1], +@z(y[1], 1@z))

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


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



The set Q consists of the following terms:

cond(TRUE, x0, x1)
cond(FALSE, x0, x1)
minus(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): MINUS(x[0], y[0]) → COND(>=@z(x[0], +@z(y[0], 1@z)), x[0], y[0])
(1): COND(TRUE, x[1], y[1]) → MINUS(x[1], +@z(y[1], 1@z))

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


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



The set Q consists of the following terms:

cond(TRUE, x0, x1)
cond(FALSE, x0, x1)
minus(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 MINUS(x, y) → COND(>=@z(x, +@z(y, 1@z)), x, y) the following chains were created:




For Pair COND(TRUE, x, y) → MINUS(x, +@z(y, 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(TRUE) = 1   
POL(MINUS(x1, x2)) = -1 + (-1)x2 + x1   
POL(+@z(x1, x2)) = x1 + x2   
POL(FALSE) = -1   
POL(COND(x1, x2, x3)) = -1 + (-1)x3 + x2   
POL(1@z) = 1   
POL(undefined) = -1   

The following pairs are in P>:

COND(TRUE, x[1], y[1]) → MINUS(x[1], +@z(y[1], 1@z))

The following pairs are in Pbound:

COND(TRUE, x[1], y[1]) → MINUS(x[1], +@z(y[1], 1@z))

The following pairs are in P:

MINUS(x[0], y[0]) → COND(>=@z(x[0], +@z(y[0], 1@z)), 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): MINUS(x[0], y[0]) → COND(>=@z(x[0], +@z(y[0], 1@z)), x[0], y[0])


The set Q consists of the following terms:

cond(TRUE, x0, x1)
cond(FALSE, x0, x1)
minus(x0, x1)


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