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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 9 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
          ↳ QDP

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

B(a, b(c(z, x, y), a)) → C(y, z, a)
C(f(c(a, y, a)), x, z) → B(y, b(x, a))
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(B(x1, x2)) = 2·x1 + x2   
POL(C(x1, x2, x3)) = x1 + 2·x2 + 2·x3   
POL(a) = 0   
POL(b(x1, x2)) = x1 + 2·x2   
POL(c(x1, x2, x3)) = 2·x1 + 2·x2 + x3   
POL(f(x1)) = 2 + x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
R is empty.
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.

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

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

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

The TRS R consists of the following rules:

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

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.


F(c(a, b(b(z, a), y), x)) → F(c(x, b(z, x), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( f(x1) ) =
/0\
\0/
+
/00\
\10/
·x1

M( a ) =
/0\
\0/

M( b(x1, x2) ) =
/1\
\1/
+
/00\
\11/
·x1+
/00\
\01/
·x2

M( c(x1, ..., x3) ) =
/1\
\0/
+
/00\
\11/
·x1+
/01\
\00/
·x2+
/01\
\10/
·x3

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


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:

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



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

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

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

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.