Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

B(b(y, z), c(a, a, a)) → F(c(z, y, z))
B(b(y, z), c(a, a, a)) → C(z, y, z)
C(y, x, f(z)) → B(z, x)
C(y, x, f(z)) → F(b(z, x))
C(y, x, f(z)) → B(f(b(z, x)), z)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

B(b(y, z), c(a, a, a)) → F(c(z, y, z))
B(b(y, z), c(a, a, a)) → C(z, y, z)
C(y, x, f(z)) → B(z, x)
C(y, x, f(z)) → F(b(z, x))
C(y, x, f(z)) → B(f(b(z, x)), z)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

B(b(y, z), c(a, a, a)) → C(z, y, z)
C(y, x, f(z)) → B(z, x)
C(y, x, f(z)) → B(f(b(z, x)), z)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


B(b(y, z), c(a, a, a)) → C(z, y, z)
The remaining pairs can at least be oriented weakly.

C(y, x, f(z)) → B(z, x)
C(y, x, f(z)) → B(f(b(z, x)), z)
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( f(x1) ) =
/0\
\0/
+
/02\
\10/
·x1

M( a ) =
/0\
\1/

M( b(x1, x2) ) =
/0\
\1/
+
/02\
\01/
·x1+
/00\
\22/
·x2

M( c(x1, ..., x3) ) =
/2\
\1/
+
/00\
\00/
·x1+
/00\
\01/
·x2+
/20\
\22/
·x3

Tuple symbols:
M( C(x1, ..., x3) ) = 0+
[0,2]
·x1+
[0,0]
·x2+
[2,1]
·x3

M( B(x1, x2) ) = 0+
[0,2]
·x1+
[0,0]
·x2


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

c(y, x, f(z)) → b(f(b(z, x)), z)
f(b(b(a, z), c(a, x, y))) → z
b(b(y, z), c(a, a, a)) → f(c(z, y, z))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof

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

C(y, x, f(z)) → B(z, x)
C(y, x, f(z)) → B(f(b(z, x)), z)

The TRS R consists of the following rules:

b(b(y, z), c(a, a, a)) → f(c(z, y, z))
f(b(b(a, z), c(a, x, y))) → z
c(y, x, f(z)) → b(f(b(z, x)), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.