YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z083.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 2 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 18 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 QDPOrderProof (⇔, 14 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPOrderProof (⇔, 162 ms)
        ↳15 QDP
        ↳16 PisEmptyProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES

(0) Obligation:

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

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

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

c(a(b(x))) → c(b(x))


(4) Obligation:

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

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

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

b(d(b(x))) → b(d(c(x)))
d(a(x)) → c(d(x))
b(b(b(x))) → c(b(a(x)))
c(d(x)) → d(b(x))
c(d(x)) → d(b(d(x)))
c(a(d(x))) → b(b(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 with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

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

The TRS R consists of the following rules:

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


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

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

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

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

(11) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(13) Obligation:

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

B(d(b(x))) → B(d(c(x)))

The TRS R consists of the following rules:

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

POL(B(x1)) = -I +
[0A,0A,-I]
·x1

POL(d(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I0A\
|0A-I0A|
\-I0A0A/
·x1

POL(b(x1)) =
/1A\
|0A|
\0A/
 +
/0A1A1A\
|0A0A-I|
\0A0A-I/
·x1

POL(c(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A0A\
|-I1A1A|
\-I0A0A/
·x1

POL(a(x1)) =
/0A\
|0A|
\1A/
 +
/0A0A0A\
|-I0A0A|
\1A1A1A/
·x1

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

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

(15) Obligation:

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

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

D(a(x)) → D(x)

The TRS R consists of the following rules:

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

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

(19) 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.

(20) Obligation:

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

D(a(x)) → D(x)

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

(21) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • D(a(x)) → D(x)
    The graph contains the following edges 1 > 1

(22) YES