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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

b(f(b(x, z)), y) → f(f(f(b(z, b(y, z)))))
c(f(f(c(x, a, z))), a, y) → b(y, f(b(a, z)))
b(b(c(b(a, a), a, z), f(a)), y) → 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(f(b(x, z)), y) → B(y, z)
B(f(b(x, z)), y) → B(z, b(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( a ) =
/1\
\0/

M( f(x1) ) =
/0\
\1/
+
/00\
\10/
·x1

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

M( c(x1, ..., x3) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2+
/00\
\11/
·x3

Tuple symbols:
M( B(x1, x2) ) = 0+
[0,1]
·x1+
[0,1]
·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:

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



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

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.