YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z123.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 18 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 19 ms)
↳4 QDP
↳5 MRRProof (⇔, 40 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 QDPOrderProof (⇔, 7 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 QDPOrderProof (⇔, 9 ms)
        ↳16 QDP
        ↳17 DependencyGraphProof (⇔, 0 ms)
        ↳18 TRUE

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(a(x1)) = 39 + x1   
POL(b(x1)) = 26 + x1   
POL(c(x1)) = 17 + x1   
POL(d(x1)) = 11 + x1   
POL(f(x1)) = 39 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(2) Obligation:

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

a(a(x)) → b(b(b(x)))
a(x) → d(c(d(x)))
b(b(b(x))) → a(f(x))
c(d(d(x))) → f(x)
f(f(x)) → f(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(x)) → B(b(b(x)))
A(a(x)) → B(b(x))
A(a(x)) → B(x)
A(x) → C(d(x))
B(b(b(x))) → A(f(x))
B(b(b(x))) → F(x)
C(d(d(x))) → F(x)
F(f(x)) → F(a(x))
F(f(x)) → A(x)

The TRS R consists of the following rules:

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

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

(5) 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:

A(a(x)) → B(b(x))
A(a(x)) → B(x)
A(x) → C(d(x))
B(b(b(x))) → F(x)
C(d(d(x))) → F(x)
F(f(x)) → A(x)


Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 2 + 2·x1   
POL(B(x1)) = 2·x1   
POL(C(x1)) = 1 + 2·x1   
POL(F(x1)) = 2·x1   
POL(a(x1)) = 3 + x1   
POL(b(x1)) = 2 + x1   
POL(c(x1)) = 3 + x1   
POL(d(x1)) = x1   
POL(f(x1)) = 3 + x1   

(6) Obligation:

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

A(a(x)) → B(b(b(x)))
B(b(b(x))) → A(f(x))
F(f(x)) → F(a(x))

The TRS R consists of the following rules:

a(a(x)) → b(b(b(x)))
a(x) → d(c(d(x)))
b(b(b(x))) → a(f(x))
c(d(d(x))) → f(x)
f(f(x)) → f(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 2 SCCs.

(8) Complex Obligation (AND)

(9) Obligation:

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

F(f(x)) → F(a(x))

The TRS R consists of the following rules:

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

Q is empty.
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.


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

POL(F(x1)) = x1   
POL(a(x1)) = 0   
POL(b(x1)) = 0   
POL(c(x1)) = 1 + x1   
POL(d(x1)) = 0   
POL(f(x1)) = 1   

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

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

(11) Obligation:

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

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

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) YES

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(15) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

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

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

f(f(x)) → f(a(x))

(16) Obligation:

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

B(b(b(x))) → A(f(x))

The TRS R consists of the following rules:

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

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(18) TRUE