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_log(TRUE, x, y) → +@z(1@z, log(/@z(-@z(x, y), y), y))
Cond_log1(TRUE, y) → 0@z
log(x, y) → Cond_log(&&(>=@z(x, 2@z), >=@z(y, 2@z)), x, y)
log(1@z, y) → Cond_log1(>=@z(y, 2@z), y)

The set Q consists of the following terms:

Cond_log(TRUE, x0, x1)
Cond_log1(TRUE, x0)
log(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_log(TRUE, x, y) → +@z(1@z, log(/@z(-@z(x, y), y), y))
Cond_log1(TRUE, y) → 0@z
log(x, y) → Cond_log(&&(>=@z(x, 2@z), >=@z(y, 2@z)), x, y)
log(1@z, y) → Cond_log1(>=@z(y, 2@z), y)

The integer pair graph contains the following rules and edges:

(0): LOG(1@z, y[0]) → COND_LOG1(>=@z(y[0], 2@z), y[0])
(1): COND_LOG(TRUE, x[1], y[1]) → LOG(/@z(-@z(x[1], y[1]), y[1]), y[1])
(2): LOG(x[2], y[2]) → COND_LOG(&&(>=@z(x[2], 2@z), >=@z(y[2], 2@z)), x[2], y[2])

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


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


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



The set Q consists of the following terms:

Cond_log(TRUE, x0, x1)
Cond_log1(TRUE, x0)
log(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
          ↳ IDependencyGraphProof

I DP problem:
The following domains are used:

z

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

(0): LOG(1@z, y[0]) → COND_LOG1(>=@z(y[0], 2@z), y[0])
(1): COND_LOG(TRUE, x[1], y[1]) → LOG(/@z(-@z(x[1], y[1]), y[1]), y[1])
(2): LOG(x[2], y[2]) → COND_LOG(&&(>=@z(x[2], 2@z), >=@z(y[2], 2@z)), x[2], y[2])

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


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


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



The set Q consists of the following terms:

Cond_log(TRUE, x0, x1)
Cond_log1(TRUE, x0)
log(x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(1): COND_LOG(TRUE, x[1], y[1]) → LOG(/@z(-@z(x[1], y[1]), y[1]), y[1])
(2): LOG(x[2], y[2]) → COND_LOG(&&(>=@z(x[2], 2@z), >=@z(y[2], 2@z)), x[2], y[2])

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


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



The set Q consists of the following terms:

Cond_log(TRUE, x0, x1)
Cond_log1(TRUE, x0)
log(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 COND_LOG(TRUE, x[1], y[1]) → LOG(/@z(-@z(x[1], y[1]), y[1]), y[1]) the following chains were created:




For Pair LOG(x[2], y[2]) → COND_LOG(&&(>=@z(x[2], 2@z), >=@z(y[2], 2@z)), x[2], y[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)) = x1 + (-1)x2   
POL(>=@z(x1, x2)) = -1   
POL(TRUE) = 1   
POL(&&(x1, x2)) = 1   
POL(2@z) = 2   
POL(LOG(x1, x2)) = -1 + x1   
POL(FALSE) = 1   
POL(COND_LOG(x1, x2, x3)) = -1 + x2 + (-1)x1   
POL(undefined) = -1   

Polynomial Interpretations with Context Sensitive Arithemetic Replacement
POL(TermCSAR-Mode @ Context)

POL(/@z(x1, y[2])1 @ {LOG_2/0}) = max{x1, (-1)x1} + (-1)min{max{x2, (-1)x2} + -1, max{x1, (-1)x1}}   

The following pairs are in P>:

LOG(x[2], y[2]) → COND_LOG(&&(>=@z(x[2], 2@z), >=@z(y[2], 2@z)), x[2], y[2])

The following pairs are in Pbound:

COND_LOG(TRUE, x[1], y[1]) → LOG(/@z(-@z(x[1], y[1]), y[1]), y[1])

The following pairs are in P:

COND_LOG(TRUE, x[1], y[1]) → LOG(/@z(-@z(x[1], y[1]), y[1]), y[1])

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(2): LOG(x[2], y[2]) → COND_LOG(&&(>=@z(x[2], 2@z), >=@z(y[2], 2@z)), x[2], y[2])


The set Q consists of the following terms:

Cond_log(TRUE, x0, x1)
Cond_log1(TRUE, x0)
log(x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(1): COND_LOG(TRUE, x[1], y[1]) → LOG(/@z(-@z(x[1], y[1]), y[1]), y[1])


The set Q consists of the following terms:

Cond_log(TRUE, x0, x1)
Cond_log1(TRUE, x0)
log(x0, x1)


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