YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-13.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 6 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 117 ms)
↳8 QDP
↳9 UsableRulesProof (⇔, 0 ms)
↳10 QDP
↳11 QDPSizeChangeProof (⇔, 1 ms)
↳12 YES

(0) Obligation:

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

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(6) Obligation:

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

C(a(x)) → C(x)
C(a(x)) → C(c(x))

The TRS R consists of the following rules:

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

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

(7) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(C(x1)) = 0A +
[0A,0A,-I]
·x1

POL(a(x1)) =
/0A\
|1A|
\-I/
 +
/0A-I0A\
|0A1A0A|
\-I0A0A/
·x1

POL(c(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A-I\
|-I0A-I|
\-I1A0A/
·x1

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

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

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

(8) Obligation:

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

C(a(x)) → C(x)

The TRS R consists of the following rules:

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

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

(9) 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.

(10) Obligation:

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

C(a(x)) → C(x)

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

(11) 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:

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

(12) YES