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-279.srs

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(A(x1)) = -I +
[0A,0A,-I]
·x1

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

POL(C(x1)) = -I +
[0A,-I,-I]
·x1

POL(B(x1)) = -I +
[0A,0A,0A]
·x1

POL(a(x1)) =
/-I\
|-I|
\0A/
+
/0A0A-I\
|0A0A-I|
\0A1A0A/
·x1

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

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

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

a(x) → x
a(b(x)) → c(b(x))
b(x) → a(a(c(x)))
c(c(x)) → b(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 2 less nodes.

(6) Obligation:

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

B(x) → C(x)
C(c(x)) → B(x)

The TRS R consists of the following rules:

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

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

(7) 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.

(8) Obligation:

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

B(x) → C(x)
C(c(x)) → B(x)

R is empty.
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:

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

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

(10) YES