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_gcd1(TRUE, x, y) → gcd(-@z(y, x), x)
Cond_gcd(TRUE, x, y) → gcd(-@z(x, y), y)
gcd(0@z, y) → y
gcd(x, 0@z) → x
gcd(x, y) → Cond_gcd(&&(>=@z(x, y), >@z(y, 0@z)), x, y)
gcd(x, y) → Cond_gcd1(&&(>@z(y, x), >@z(x, 0@z)), x, y)

The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(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_gcd1(TRUE, x, y) → gcd(-@z(y, x), x)
Cond_gcd(TRUE, x, y) → gcd(-@z(x, y), y)
gcd(0@z, y) → y
gcd(x, 0@z) → x
gcd(x, y) → Cond_gcd(&&(>=@z(x, y), >@z(y, 0@z)), x, y)
gcd(x, y) → Cond_gcd1(&&(>@z(y, x), >@z(x, 0@z)), x, y)

The integer pair graph contains the following rules and edges:

(0): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])
(1): GCD(x[1], y[1]) → COND_GCD(&&(>=@z(x[1], y[1]), >@z(y[1], 0@z)), x[1], y[1])
(2): COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
(3): COND_GCD(TRUE, x[3], y[3]) → GCD(-@z(x[3], y[3]), y[3])

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


(1) -> (3), if ((x[1]* x[3])∧(y[1]* y[3])∧(&&(>=@z(x[1], y[1]), >@z(y[1], 0@z)) →* TRUE))


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


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


(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], y[3]) →* x[0]))


(3) -> (1), if ((y[3]* y[1])∧(-@z(x[3], y[3]) →* x[1]))



The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(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): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])
(1): GCD(x[1], y[1]) → COND_GCD(&&(>=@z(x[1], y[1]), >@z(y[1], 0@z)), x[1], y[1])
(2): COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
(3): COND_GCD(TRUE, x[3], y[3]) → GCD(-@z(x[3], y[3]), y[3])

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


(1) -> (3), if ((x[1]* x[3])∧(y[1]* y[3])∧(&&(>=@z(x[1], y[1]), >@z(y[1], 0@z)) →* TRUE))


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


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


(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], y[3]) →* x[0]))


(3) -> (1), if ((y[3]* y[1])∧(-@z(x[3], y[3]) →* x[1]))



The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(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 GCD(x, y) → COND_GCD1(&&(>@z(y, x), >@z(x, 0@z)), x, y) the following chains were created:




For Pair GCD(x, y) → COND_GCD(&&(>=@z(x, y), >@z(y, 0@z)), x, y) the following chains were created:




For Pair COND_GCD1(TRUE, x, y) → GCD(-@z(y, x), x) the following chains were created:




For Pair COND_GCD(TRUE, x, y) → GCD(-@z(x, y), 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 + (-1)x2   
POL(COND_GCD1(x1, x2, x3)) = -1 + x3 + x2   
POL(>=@z(x1, x2)) = -1   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = -1   
POL(COND_GCD(x1, x2, x3)) = -1 + x3 + x2 + x1   
POL(GCD(x1, x2)) = -1 + x2 + x1   
POL(FALSE) = -1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

GCD(x[1], y[1]) → COND_GCD(&&(>=@z(x[1], y[1]), >@z(y[1], 0@z)), x[1], y[1])

The following pairs are in Pbound:

COND_GCD(TRUE, x[3], y[3]) → GCD(-@z(x[3], y[3]), y[3])

The following pairs are in P:

GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])
COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
COND_GCD(TRUE, x[3], y[3]) → GCD(-@z(x[3], y[3]), y[3])

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

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


↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(0): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])
(1): GCD(x[1], y[1]) → COND_GCD(&&(>=@z(x[1], y[1]), >@z(y[1], 0@z)), x[1], y[1])
(2): COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])

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


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


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



The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(x0, x1)


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

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

I DP problem:
The following domains are used:

z

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

(2): COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
(0): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])

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


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



The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(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_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2]) the following chains were created:




For Pair GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0]) 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(COND_GCD1(x1, x2, x3)) = -1 + (2)x3 + (2)x2   
POL(0@z) = 0   
POL(TRUE) = 2   
POL(&&(x1, x2)) = 0   
POL(GCD(x1, x2)) = (2)x2 + (2)x1   
POL(FALSE) = 0   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])

The following pairs are in Pbound:

COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])

The following pairs are in P:
none

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

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


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

I DP problem:
The following domains are used:

z

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

(0): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])


The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(x0, x1)


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

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

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:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ 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:

(0): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])
(2): COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
(3): COND_GCD(TRUE, x[3], y[3]) → GCD(-@z(x[3], y[3]), y[3])

(3) -> (0), if ((y[3]* y[0])∧(-@z(x[3], y[3]) →* x[0]))


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


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



The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ 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:

(2): COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
(0): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])

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


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



The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(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_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2]) the following chains were created:




For Pair GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0]) 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(COND_GCD1(x1, x2, x3)) = -1 + (2)x3 + (2)x2   
POL(0@z) = 0   
POL(TRUE) = -1   
POL(&&(x1, x2)) = 2   
POL(GCD(x1, x2)) = (2)x2 + (2)x1   
POL(FALSE) = -1   
POL(undefined) = -1   
POL(>@z(x1, x2)) = -1   

The following pairs are in P>:

COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])
GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])

The following pairs are in Pbound:

COND_GCD1(TRUE, x[2], y[2]) → GCD(-@z(y[2], x[2]), x[2])

The following pairs are in P:
none

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

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


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

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:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(x0, x1)


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

↳ ITRS
  ↳ ITRStoIDPProof
    ↳ IDP
      ↳ UsableRulesProof
        ↳ IDP
          ↳ IDPNonInfProof
            ↳ AND
              ↳ IDP
              ↳ 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:

(0): GCD(x[0], y[0]) → COND_GCD1(&&(>@z(y[0], x[0]), >@z(x[0], 0@z)), x[0], y[0])


The set Q consists of the following terms:

Cond_gcd1(TRUE, x0, x1)
Cond_gcd(TRUE, x0, x1)
gcd(x0, x1)


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