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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 MRRProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 MRRProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE

(0) Obligation:

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

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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:

F(a, f(a, f(a, f(x, b)))) → F(f(a, f(a, f(a, x))), b)
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, f(a, x)))
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, x))
F(a, f(a, f(a, f(x, b)))) → F(a, x)
F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)
F(f(f(a, x), b), b) → F(a, f(f(x, b), b))
F(f(f(a, x), b), b) → F(f(x, b), b)
F(f(f(a, x), b), b) → F(x, b)

The TRS R consists of the following rules:

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(f(f(a, x), b), b) → F(f(x, b), b)
F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)
F(f(f(a, x), b), b) → F(x, b)

The TRS R consists of the following rules:

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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

(6) 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:

F(f(f(a, x), b), b) → F(f(x, b), b)
F(f(f(a, x), b), b) → F(x, b)


Used ordering: Polynomial interpretation [POLO]:

POL(F(x1, x2)) = x1 + x2   
POL(a) = 1   
POL(b) = 0   
POL(f(x1, x2)) = x1 + x2   

(7) Obligation:

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

F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)

The TRS R consists of the following rules:

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(F(x1, x2)) =
/0\
\0/
 +
/10\
\10/
·x1 +
/00\
\00/
·x2

POL(f(x1, x2)) =
/1\
\0/
 +
/00\
\00/
·x1 +
/01\
\10/
·x2

POL(a) =
/0\
\0/

POL(b) =
/0\
\1/

The following usable rules [FROCOS05] were oriented:

f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)

(9) Obligation:

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

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

F(a, f(a, f(a, f(x, b)))) → F(a, f(a, x))
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, f(a, x)))
F(a, f(a, f(a, f(x, b)))) → F(a, x)

The TRS R consists of the following rules:

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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

(13) 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:

F(a, f(a, f(a, f(x, b)))) → F(a, f(a, x))
F(a, f(a, f(a, f(x, b)))) → F(a, x)


Used ordering: Polynomial interpretation [POLO]:

POL(F(x1, x2)) = x1 + x2   
POL(a) = 1   
POL(b) = 0   
POL(f(x1, x2)) = x1 + x2   

(14) Obligation:

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

F(a, f(a, f(a, f(x, b)))) → F(a, f(a, f(a, x)))

The TRS R consists of the following rules:

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(a, f(a, f(a, f(x, b)))) → F(a, f(a, f(a, x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(F(x1, x2)) =
/0A\
|-I|
\2A/
 +
/-I3A-I\
|-I-I-I|
\-I-I2A/
·x1 +
/-I-I3A\
|-I-I-I|
\-I-I0A/
·x2

POL(a) =
/0A\
|2A|
\-I/

POL(f(x1, x2)) =
/-I\
|-I|
\-I/
 +
/1A-I-I\
|-I-I0A|
\-I-I-I/
·x1 +
/-I-I0A\
|0A-I-I|
\-I0A-I/
·x2

POL(b) =
/0A\
|0A|
\3A/

The following usable rules [FROCOS05] were oriented:

f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)

(16) Obligation:

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

f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE