Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, a(x, y)) → A(1, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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(0, a(1, a(x, y))) → B(1, a(0, a(x, y)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(x, y)) → A(1, a(1, a(x, y)))
A(0, b(0, x)) → A(0, x)
A(0, a(1, a(x, y))) → A(1, a(0, a(x, y)))
A(0, a(x, y)) → A(1, a(x, y))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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 4 less nodes.

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

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))
A(0, b(0, x)) → A(0, x)

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, x)
A(0, b(0, x)) → A(0, x)


Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1   
POL(1) = 0   
POL(A(x1, x2)) = 2 + x1 + x2   
POL(B(x1, x2)) = 1 + x1 + x2   
POL(a(x1, x2)) = 1 + x1 + x2   
POL(b(x1, x2)) = x1 + x2   



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

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, x) → B(0, b(0, x))
A(0, b(0, x)) → B(0, a(0, x))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(0, x) → B(0, b(0, x)) at position [1] we obtained the following new rules:

A(0, a(1, a(x0, x1))) → B(0, b(1, a(0, a(x0, x1))))



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

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → B(0, a(0, x))
A(0, a(1, a(x0, x1))) → B(0, b(1, a(0, a(x0, x1))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing

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

B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, x)) → B(0, a(0, x))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(0, b(0, x)) → B(0, a(0, x)) at position [1] we obtained the following new rules:

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, a(x0, x1))) → B(0, a(1, a(1, a(x0, x1))))



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

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
B(0, a(1, a(x, y))) → A(0, a(x, y))
A(0, b(0, a(x0, x1))) → B(0, a(1, a(1, a(x0, x1))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(0, a(1, a(x, y))) → A(0, a(x, y)) at position [1] we obtained the following new rules:

B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))
B(0, a(1, a(0, a(x0, x1)))) → A(0, a(1, a(1, a(x0, x1))))



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

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, a(x0, x1))) → B(0, a(1, a(1, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))
B(0, a(1, a(0, a(x0, x1)))) → A(0, a(1, a(1, a(x0, x1))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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 3 less nodes.

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

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
A(0, b(0, x0)) → B(0, b(0, b(0, x0)))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule A(0, b(0, x0)) → B(0, b(0, b(0, x0))) at position [1] we obtained the following new rules:

A(0, b(0, a(1, a(x0, x1)))) → B(0, b(0, b(1, a(0, a(x0, x1)))))



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

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))
A(0, b(0, a(1, a(x0, x1)))) → B(0, b(0, b(1, a(0, a(x0, x1)))))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ QDPOrderProof

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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.


A(0, b(0, b(0, x0))) → B(0, b(0, a(0, x0)))
The remaining pairs can at least be oriented weakly.

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( b(x1, x2) ) =
/0\
\0/
+
/00\
\00/
·x1+
/00\
\00/
·x2

M( a(x1, x2) ) =
/0\
\1/
+
/01\
\00/
·x1+
/00\
\10/
·x2

M( 0 ) =
/0\
\0/

M( 1 ) =
/0\
\0/

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

M( A(x1, x2) ) = 1+
[0,0]
·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:

a(0, x) → b(0, b(0, x))
a(0, b(0, x)) → b(0, a(0, x))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))



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

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0)))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule B(0, a(1, a(0, x0))) → A(0, b(0, b(0, x0))) at position [1] we obtained the following new rules:

B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, b(0, b(1, a(0, a(x0, x1)))))



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

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, a(1, a(x0, x1))))) → A(0, b(0, b(1, a(0, a(x0, x1)))))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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
            ↳ QDP
              ↳ Narrowing
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ Narrowing
                        ↳ QDP
                          ↳ Narrowing
                            ↳ QDP
                              ↳ DependencyGraphProof
                                ↳ QDP
                                  ↳ Narrowing
                                    ↳ QDP
                                      ↳ DependencyGraphProof
                                        ↳ QDP
                                          ↳ QDPOrderProof
                                            ↳ QDP
                                              ↳ Narrowing
                                                ↳ QDP
                                                  ↳ DependencyGraphProof
                                                    ↳ QDP

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

A(0, b(0, a(1, a(x0, x1)))) → B(0, a(1, a(0, a(x0, x1))))
B(0, a(1, a(0, b(0, x0)))) → A(0, b(0, a(0, x0)))

The TRS R consists of the following rules:

a(0, b(0, x)) → b(0, a(0, x))
a(0, x) → b(0, b(0, x))
a(0, a(1, a(x, y))) → a(1, a(0, a(x, y)))
b(0, a(1, a(x, y))) → b(1, a(0, a(x, y)))
a(0, a(x, y)) → a(1, a(1, a(x, y)))

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