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:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

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:

A(0, x) → C(c(x))
C(c(b(c(x)))) → A(0, c(x))
A(0, x) → C(x)
C(c(x)) → C(b(c(x)))

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

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:

A(0, x) → C(c(x))
C(c(b(c(x)))) → A(0, c(x))
A(0, x) → C(x)
C(c(x)) → C(b(c(x)))

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

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 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof

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

A(0, x) → C(c(x))
C(c(b(c(x)))) → A(0, c(x))
A(0, x) → C(x)

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A(0, x) → C(x)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 2   
POL(A(x1, x2)) = 2·x1 + 2·x2   
POL(C(x1)) = 2·x1   
POL(a(x1, x2)) = 2·x1 + x2   
POL(b(x1)) = x1   
POL(c(x1)) = 2 + x1   



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

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

A(0, x) → C(c(x))
C(c(b(c(x)))) → A(0, c(x))

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

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.


C(c(b(c(x)))) → A(0, c(x))
The remaining pairs can at least be oriented weakly.

A(0, x) → C(c(x))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( c(x1) ) =
/1\
\0/
+
/01\
\10/
·x1

M( a(x1, x2) ) =
/1\
\0/
+
/01\
\01/
·x1+
/10\
\01/
·x2

M( b(x1) ) =
/0\
\1/
+
/00\
\01/
·x1

M( 0 ) =
/0\
\1/

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

M( A(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:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))



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

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

A(0, x) → C(c(x))

The TRS R consists of the following rules:

c(c(b(c(x)))) → b(a(0, c(x)))
c(c(x)) → b(c(b(c(x))))
a(0, x) → c(c(x))

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 1 less node.