YES Termination w.r.t. Q proof of SK90_4.32.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 75 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 QTRSRRRProof (⇔, 0 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 15 ms)
↳8 QDP
↳9 QDPOrderProof (⇔, 21 ms)
↳10 QDP
↳11 DependencyGraphProof (⇔, 0 ms)
↳12 AND
    ↳13 QDP
        ↳14 QDPOrderProof (⇔, 6 ms)
        ↳15 QDP
        ↳16 PisEmptyProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 QDPOrderProof (⇔, 4 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES

(0) Obligation:

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

a(b(x)) → b(a(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(c(x)))
u(a(x)) → x
v(b(x)) → x
w(c(x)) → x
a(u(x)) → x
b(v(x)) → x
c(w(x)) → x

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(u(x1)) = 1 + x1   
POL(v(x1)) = x1   
POL(w(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

u(a(x)) → x
a(u(x)) → x


(2) Obligation:

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

a(b(x)) → b(a(a(x)))
b(c(x)) → c(b(b(x)))
c(a(x)) → a(c(c(x)))
v(b(x)) → x
w(c(x)) → x
b(v(x)) → x
c(w(x)) → x

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(v(x1)) = 1 + x1   
POL(w(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

v(b(x)) → x
b(v(x)) → x


(4) Obligation:

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

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

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(w(x1)) = 1 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

w(c(x)) → x
c(w(x)) → x


(6) Obligation:

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

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

Q is empty.

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) Obligation:

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

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

The TRS R consists of the following rules:

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


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

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

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

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

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.

(12) Complex Obligation (AND)

(13) Obligation:

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

B(c(x)) → B(x)
B(c(x)) → B(b(x))

The TRS R consists of the following rules:

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

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

(14) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

(15) Obligation:

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

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

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) YES

(18) Obligation:

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

A(b(x)) → A(x)
A(b(x)) → A(a(x))

The TRS R consists of the following rules:

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

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

(19) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


A(b(x)) → A(x)
A(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)) = x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1   

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

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

(20) Obligation:

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

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

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) YES