YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/beans6.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 24 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 47 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 51 ms)
↳10 QDP
↳11 PisEmptyProof (⇔, 0 ms)
↳12 YES

(0) Obligation:

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

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

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

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

The TRS R consists of the following rules:

a(a(b(x))) → c(b(a(x)))
a(c(x)) → c(a(x))
b(c(x)) → a(b(x))
b(a(a(x))) → a(b(d(x)))
d(a(x)) → a(d(x))
d(b(x)) → b(a(x))
a(a(x)) → a(b(a(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.


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

POL(A(x1)) = x1   
POL(B(x1)) = x1   
POL(D(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = 1 + x1   
POL(d(x1)) = 1 + x1   

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

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

A(a(x)) → A(b(a(x)))

The TRS R consists of the following rules:

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

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

(9) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(A(x1)) = [4]x1   
POL(a(x1)) = [2] + [2]x1   
POL(b(x1)) = [1/2]x1   
POL(c(x1)) = [4] + [2]x1   
POL(d(x1)) = [1] + [2]x1   
The value of delta used in the strict ordering is 4.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(10) Obligation:

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

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

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) YES