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-65.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 QDP
↳7 QDPOrderProof (⇔, 134 ms)
↳8 QDP
↳9 QDPSizeChangeProof (⇔, 0 ms)
↳10 YES

(0) Obligation:

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

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

The TRS R consists of the following rules:

a(x) → x
a(a(x)) → b(a(c(b(a(x)))))
b(c(x)) → c(a(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 1 less node.

(6) Obligation:

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

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

The TRS R consists of the following rules:

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


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

POL(B(x1)) = 0A +
[-I,0A,-I]
·x1

POL(c(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|0A-I-I|
\0A0A0A/
·x1

POL(A(x1)) = 0A +
[0A,-I,-I]
·x1

POL(a(x1)) =
/1A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A-I|
\0A0A0A/
·x1

POL(b(x1)) =
/0A\
|1A|
\1A/
 +
/-I-I-I\
|0A0A0A|
\0A0A0A/
·x1

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

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

(8) Obligation:

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

B(c(x)) → A(x)
A(a(x)) → B(a(x))

The TRS R consists of the following rules:

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

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

(9) 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:

  • A(a(x)) → B(a(x))
    The graph contains the following edges 1 >= 1

  • B(c(x)) → A(x)
    The graph contains the following edges 1 > 1

(10) YES