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

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 19 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 UsableRulesProof (⇔, 0 ms)
↳10 QDP
↳11 QDPOrderProof (⇔, 0 ms)
↳12 QDP
↳13 PisEmptyProof (⇔, 0 ms)
↳14 YES

(0) Obligation:

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

1(q0(1(x))) → 0(1(q1(x)))
1(q0(0(x))) → 0(0(q1(x)))
1(q1(1(x))) → 1(1(q1(x)))
1(q1(0(x))) → 1(0(q1(x)))
0(q1(x)) → q2(1(x))
1(q2(x)) → q2(1(x))
0(q2(x)) → 0(q0(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:

11(q0(1(x))) → 01(1(q1(x)))
11(q0(1(x))) → 11(q1(x))
11(q0(0(x))) → 01(0(q1(x)))
11(q0(0(x))) → 01(q1(x))
11(q1(1(x))) → 11(1(q1(x)))
11(q1(1(x))) → 11(q1(x))
11(q1(0(x))) → 11(0(q1(x)))
11(q1(0(x))) → 01(q1(x))
01(q1(x)) → 11(x)
11(q2(x)) → 11(x)
01(q2(x)) → 01(q0(x))

The TRS R consists of the following rules:

1(q0(1(x))) → 0(1(q1(x)))
1(q0(0(x))) → 0(0(q1(x)))
1(q1(1(x))) → 1(1(q1(x)))
1(q1(0(x))) → 1(0(q1(x)))
0(q1(x)) → q2(1(x))
1(q2(x)) → q2(1(x))
0(q2(x)) → 0(q0(x))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

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

11(q0(1(x))) → 11(q1(x))
11(q1(1(x))) → 11(1(q1(x)))
11(q2(x)) → 11(x)
11(q0(0(x))) → 01(q1(x))
01(q1(x)) → 11(x)
11(q1(1(x))) → 11(q1(x))
11(q1(0(x))) → 11(0(q1(x)))
11(q1(0(x))) → 01(q1(x))

The TRS R consists of the following rules:

1(q0(1(x))) → 0(1(q1(x)))
1(q0(0(x))) → 0(0(q1(x)))
1(q1(1(x))) → 1(1(q1(x)))
1(q1(0(x))) → 1(0(q1(x)))
0(q1(x)) → q2(1(x))
1(q2(x)) → q2(1(x))
0(q2(x)) → 0(q0(x))

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

(5) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


11(q0(1(x))) → 11(q1(x))
11(q2(x)) → 11(x)
11(q0(0(x))) → 01(q1(x))
11(q1(0(x))) → 01(q1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(01(x1)) = x1   
POL(1(x1)) = x1   
POL(11(x1)) = x1   
POL(q0(x1)) = 1 + x1   
POL(q1(x1)) = x1   
POL(q2(x1)) = 1 + x1   

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

1(q1(0(x))) → 1(0(q1(x)))
1(q1(1(x))) → 1(1(q1(x)))
0(q1(x)) → q2(1(x))
1(q0(1(x))) → 0(1(q1(x)))
1(q0(0(x))) → 0(0(q1(x)))
1(q2(x)) → q2(1(x))
0(q2(x)) → 0(q0(x))

(6) Obligation:

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

11(q1(1(x))) → 11(1(q1(x)))
01(q1(x)) → 11(x)
11(q1(1(x))) → 11(q1(x))
11(q1(0(x))) → 11(0(q1(x)))

The TRS R consists of the following rules:

1(q0(1(x))) → 0(1(q1(x)))
1(q0(0(x))) → 0(0(q1(x)))
1(q1(1(x))) → 1(1(q1(x)))
1(q1(0(x))) → 1(0(q1(x)))
0(q1(x)) → q2(1(x))
1(q2(x)) → q2(1(x))
0(q2(x)) → 0(q0(x))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(8) Obligation:

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

11(q1(1(x))) → 11(q1(x))

The TRS R consists of the following rules:

1(q0(1(x))) → 0(1(q1(x)))
1(q0(0(x))) → 0(0(q1(x)))
1(q1(1(x))) → 1(1(q1(x)))
1(q1(0(x))) → 1(0(q1(x)))
0(q1(x)) → q2(1(x))
1(q2(x)) → q2(1(x))
0(q2(x)) → 0(q0(x))

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

(9) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(10) Obligation:

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

11(q1(1(x))) → 11(q1(x))

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

(11) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


11(q1(1(x))) → 11(q1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(1(x1)) = 1 + x1   
POL(11(x1)) = x1   
POL(q1(x1)) = x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(12) Obligation:

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) YES