YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Gebhardt_06/14.srs

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔, 10 ms)
↳2 QDP
↳3 QDPOrderProof (⇔, 41 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 252 ms)
↳8 QDP
↳9 PisEmptyProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

0(0(0(0(x)))) → 0(1(1(1(x))))
1(1(0(1(x)))) → 0(0(0(0(x))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

01(0(0(0(x)))) → 01(1(1(1(x))))
01(0(0(0(x)))) → 11(1(1(x)))
01(0(0(0(x)))) → 11(1(x))
01(0(0(0(x)))) → 11(x)
11(1(0(1(x)))) → 01(0(0(0(x))))
11(1(0(1(x)))) → 01(0(0(x)))
11(1(0(1(x)))) → 01(0(x))
11(1(0(1(x)))) → 01(x)

The TRS R consists of the following rules:

0(0(0(0(x)))) → 0(1(1(1(x))))
1(1(0(1(x)))) → 0(0(0(0(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


01(0(0(0(x)))) → 11(1(1(x)))
01(0(0(0(x)))) → 11(1(x))
01(0(0(0(x)))) → 11(x)
11(1(0(1(x)))) → 01(0(0(x)))
11(1(0(1(x)))) → 01(0(x))
11(1(0(1(x)))) → 01(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(01(x1)) = 1 + x1   
POL(1(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

1(1(0(1(x)))) → 0(0(0(0(x))))
0(0(0(0(x)))) → 0(1(1(1(x))))

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

01(0(0(0(x)))) → 01(1(1(1(x))))
11(1(0(1(x)))) → 01(0(0(0(x))))

The TRS R consists of the following rules:

0(0(0(0(x)))) → 0(1(1(1(x))))
1(1(0(1(x)))) → 0(0(0(0(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

01(0(0(0(x)))) → 01(1(1(1(x))))

The TRS R consists of the following rules:

0(0(0(0(x)))) → 0(1(1(1(x))))
1(1(0(1(x)))) → 0(0(0(0(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


01(0(0(0(x)))) → 01(1(1(1(x))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(01(x1)) = -I +
[0A,-I,-I]
·x1

POL(0(x1)) =
/0A\
|0A|
\0A/
 +
/-I1A0A\
|0A-I0A|
\-I0A-I/
·x1

POL(1(x1)) =
/0A\
|1A|
\0A/
 +
/-I-I0A\
|1A0A1A|
\0A-I-I/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

1(1(0(1(x)))) → 0(0(0(0(x))))
0(0(0(0(x)))) → 0(1(1(1(x))))

(8) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

0(0(0(0(x)))) → 0(1(1(1(x))))
1(1(0(1(x)))) → 0(0(0(0(x))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) YES