YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Gebhardt_06/04.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 MRRProof (⇔, 81 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 289 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 TRUE

(0) Obligation:

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

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

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

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

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 2 + x1   
POL(01(x1)) = 2 + x1   
POL(1(x1)) = 2 + x1   
POL(11(x1)) = 2 + x1   

(6) Obligation:

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

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

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(0(1(0(x))))
1(0(1(0(x)))) → 0(1(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)))) → 11(0(1(0(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,0A,0A]
·x1

POL(0(x1)) =
/0A\
|0A|
\1A/
 +
/-I0A-I\
|-I-I0A|
\0A-I1A/
·x1

POL(11(x1)) = -I +
[1A,0A,-I]
·x1

POL(1(x1)) =
/0A\
|0A|
\0A/
 +
/1A0A-I\
|-I0A0A|
\0A-I-I/
·x1

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

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

(8) Obligation:

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

11(0(1(0(x)))) → 01(1(0(0(x))))

The TRS R consists of the following rules:

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

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) TRUE