YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Trafo_06/dup06.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 25 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 MNOCProof (⇔, 1 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 8 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 12 ms)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔, 0 ms)
        ↳15 YES
    ↳16 QDP
        ↳17 QDPOrderProof (⇔, 30 ms)
        ↳18 QDP
        ↳19 PisEmptyProof (⇔, 0 ms)
        ↳20 YES

(0) Obligation:

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

a(a(b(b(x)))) → b(b(c(c(a(a(x))))))
b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(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:

A(a(b(b(x)))) → B(b(c(c(a(a(x))))))
A(a(b(b(x)))) → B(c(c(a(a(x)))))
A(a(b(b(x)))) → A(a(x))
A(a(b(b(x)))) → A(x)
B(b(c(c(x)))) → B(b(b(b(x))))
B(b(c(c(x)))) → B(b(b(x)))
B(b(c(c(x)))) → B(b(x))
B(b(c(c(x)))) → B(x)
B(b(a(a(x)))) → A(a(c(c(b(b(x))))))
B(b(a(a(x)))) → A(c(c(b(b(x)))))
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)

The TRS R consists of the following rules:

a(a(b(b(x)))) → b(b(c(c(a(a(x))))))
b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(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 2 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

B(b(c(c(x)))) → B(b(b(x)))
B(b(c(c(x)))) → B(b(b(b(x))))
B(b(c(c(x)))) → B(b(x))
B(b(c(c(x)))) → B(x)
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)

The TRS R consists of the following rules:

a(a(b(b(x)))) → b(b(c(c(a(a(x))))))
b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

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

(6) 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.

(7) Obligation:

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

B(b(c(c(x)))) → B(b(b(x)))
B(b(c(c(x)))) → B(b(b(b(x))))
B(b(c(c(x)))) → B(b(x))
B(b(c(c(x)))) → B(x)
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)

The TRS R consists of the following rules:

b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

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

(8) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(9) Obligation:

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

B(b(c(c(x)))) → B(b(b(x)))
B(b(c(c(x)))) → B(b(b(b(x))))
B(b(c(c(x)))) → B(b(x))
B(b(c(c(x)))) → B(x)
B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)

The TRS R consists of the following rules:

b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

The set Q consists of the following terms:

b(b(c(c(x0))))
b(b(a(a(x0))))

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(b(a(a(x)))) → B(b(x))
B(b(a(a(x)))) → B(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(B(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   

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

b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

(11) Obligation:

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

B(b(c(c(x)))) → B(b(b(x)))
B(b(c(c(x)))) → B(b(b(b(x))))
B(b(c(c(x)))) → B(b(x))
B(b(c(c(x)))) → B(x)

The TRS R consists of the following rules:

b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

The set Q consists of the following terms:

b(b(c(c(x0))))
b(b(a(a(x0))))

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

(12) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(b(c(c(x)))) → B(b(b(x)))
B(b(c(c(x)))) → B(b(b(b(x))))
B(b(c(c(x)))) → B(b(x))
B(b(c(c(x)))) → B(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(B(x1)) = x1   
POL(a(x1)) = 1   
POL(b(x1)) = x1   
POL(c(x1)) = 1 + x1   

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

b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

(13) Obligation:

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

b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

The set Q consists of the following terms:

b(b(c(c(x0))))
b(b(a(a(x0))))

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) YES

(16) Obligation:

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

A(a(b(b(x)))) → A(x)
A(a(b(b(x)))) → A(a(x))

The TRS R consists of the following rules:

a(a(b(b(x)))) → b(b(c(c(a(a(x))))))
b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

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

(17) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


A(a(b(b(x)))) → A(x)
A(a(b(b(x)))) → A(a(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 0   

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

a(a(b(b(x)))) → b(b(c(c(a(a(x))))))
b(b(c(c(x)))) → c(c(b(b(b(b(x))))))

(18) Obligation:

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

a(a(b(b(x)))) → b(b(c(c(a(a(x))))))
b(b(c(c(x)))) → c(c(b(b(b(b(x))))))
b(b(a(a(x)))) → a(a(c(c(b(b(x))))))

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES