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:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

+1(j(x), j(y)) → +1(x, y)
*1(*(x, y), z) → *1(y, z)
OPP(j(x)) → OPP(x)
+1(1(x), j(y)) → +1(x, y)
+1(j(x), 1(y)) → +1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
-1(x, y) → +1(x, opp(y))
+1(j(x), 1(y)) → 01(+(x, y))
+1(1(x), j(y)) → 01(+(x, y))
+1(0(x), 0(y)) → +1(x, y)
-1(x, y) → OPP(y)
*1(j(x), y) → *1(x, y)
OPP(0(x)) → OPP(x)
+1(+(x, y), z) → +1(y, z)
*1(1(x), y) → 01(*(x, y))
*1(1(x), y) → *1(x, y)
+1(j(x), j(y)) → +1(+(x, y), j(#))
+1(0(x), 0(y)) → 01(+(x, y))
*1(j(x), y) → 01(*(x, y))
*1(1(x), y) → +1(0(*(x, y)), y)
*1(j(x), y) → -1(0(*(x, y)), y)
+1(1(x), 1(y)) → +1(x, y)
OPP(0(x)) → 01(opp(x))
+1(j(x), 0(y)) → +1(x, y)
+1(0(x), j(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
OPP(1(x)) → OPP(x)
*1(0(x), y) → 01(*(x, y))
*1(0(x), y) → *1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ EdgeDeletionProof

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

+1(j(x), j(y)) → +1(x, y)
*1(*(x, y), z) → *1(y, z)
OPP(j(x)) → OPP(x)
+1(1(x), j(y)) → +1(x, y)
+1(j(x), 1(y)) → +1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
-1(x, y) → +1(x, opp(y))
+1(j(x), 1(y)) → 01(+(x, y))
+1(1(x), j(y)) → 01(+(x, y))
+1(0(x), 0(y)) → +1(x, y)
-1(x, y) → OPP(y)
*1(j(x), y) → *1(x, y)
OPP(0(x)) → OPP(x)
+1(+(x, y), z) → +1(y, z)
*1(1(x), y) → 01(*(x, y))
*1(1(x), y) → *1(x, y)
+1(j(x), j(y)) → +1(+(x, y), j(#))
+1(0(x), 0(y)) → 01(+(x, y))
*1(j(x), y) → 01(*(x, y))
*1(1(x), y) → +1(0(*(x, y)), y)
*1(j(x), y) → -1(0(*(x, y)), y)
+1(1(x), 1(y)) → +1(x, y)
OPP(0(x)) → 01(opp(x))
+1(j(x), 0(y)) → +1(x, y)
+1(0(x), j(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
OPP(1(x)) → OPP(x)
*1(0(x), y) → 01(*(x, y))
*1(0(x), y) → *1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
QDP
          ↳ DependencyGraphProof

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

+1(j(x), j(y)) → +1(x, y)
*1(*(x, y), z) → *1(y, z)
OPP(j(x)) → OPP(x)
+1(j(x), 1(y)) → +1(x, y)
+1(1(x), j(y)) → +1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
-1(x, y) → +1(x, opp(y))
+1(0(x), 0(y)) → +1(x, y)
+1(1(x), j(y)) → 01(+(x, y))
+1(j(x), 1(y)) → 01(+(x, y))
-1(x, y) → OPP(y)
*1(j(x), y) → *1(x, y)
OPP(0(x)) → OPP(x)
+1(+(x, y), z) → +1(y, z)
*1(1(x), y) → 01(*(x, y))
*1(1(x), y) → *1(x, y)
+1(j(x), j(y)) → +1(+(x, y), j(#))
+1(0(x), 0(y)) → 01(+(x, y))
*1(1(x), y) → +1(0(*(x, y)), y)
*1(j(x), y) → 01(*(x, y))
+1(1(x), 1(y)) → +1(x, y)
*1(j(x), y) → -1(0(*(x, y)), y)
OPP(0(x)) → 01(opp(x))
+1(+(x, y), z) → +1(x, +(y, z))
+1(0(x), j(y)) → +1(x, y)
+1(j(x), 0(y)) → +1(x, y)
OPP(1(x)) → OPP(x)
*1(0(x), y) → 01(*(x, y))
*1(0(x), y) → *1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 11 less nodes.

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

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

OPP(0(x)) → OPP(x)
OPP(1(x)) → OPP(x)
OPP(j(x)) → OPP(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


OPP(0(x)) → OPP(x)
The remaining pairs can at least be oriented weakly.

OPP(1(x)) → OPP(x)
OPP(j(x)) → OPP(x)
Used ordering: Combined order from the following AFS and order.
OPP(x1)  =  OPP(x1)
0(x1)  =  0(x1)
1(x1)  =  x1
j(x1)  =  x1

Lexicographic path order with status [19].
Quasi-Precedence:
trivial

Status:
OPP1: [1]
01: [1]


The following usable rules [14] were oriented: none



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

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

OPP(1(x)) → OPP(x)
OPP(j(x)) → OPP(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


OPP(1(x)) → OPP(x)
The remaining pairs can at least be oriented weakly.

OPP(j(x)) → OPP(x)
Used ordering: Combined order from the following AFS and order.
OPP(x1)  =  OPP(x1)
1(x1)  =  1(x1)
j(x1)  =  x1

Lexicographic path order with status [19].
Quasi-Precedence:
[OPP1, 11]

Status:
OPP1: [1]
11: [1]


The following usable rules [14] were oriented: none



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

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

OPP(j(x)) → OPP(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


OPP(j(x)) → OPP(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
OPP(x1)  =  OPP(x1)
j(x1)  =  j(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
[OPP1, j1]

Status:
OPP1: [1]
j1: [1]


The following usable rules [14] were oriented: none



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

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP
              ↳ QDP

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

+1(j(x), j(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(x, y)
+1(1(x), j(y)) → +1(x, y)
+1(j(x), 1(y)) → +1(x, y)
+1(0(x), j(y)) → +1(x, y)
+1(j(x), 0(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(0(x), 0(y)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)
+1(j(x), j(y)) → +1(+(x, y), j(#))
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

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

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

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

*1(*(x, y), z) → *1(y, z)
*1(j(x), y) → *1(x, y)
*1(1(x), y) → *1(x, y)
*1(0(x), y) → *1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*1(*(x, y), z) → *1(y, z)
*1(j(x), y) → *1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
The remaining pairs can at least be oriented weakly.

*1(1(x), y) → *1(x, y)
*1(0(x), y) → *1(x, y)
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1)
*(x1, x2)  =  *(x1, x2)
j(x1)  =  j(x1)
1(x1)  =  x1
0(x1)  =  x1
-(x1, x2)  =  -(x1, x2)
+(x1, x2)  =  +(x1, x2)
opp(x1)  =  opp(x1)
#  =  #

Lexicographic path order with status [19].
Quasi-Precedence:
*^11 > [*2, -2]
+2 > j1 > [*2, -2]
opp1 > j1 > [*2, -2]
# > [*2, -2]

Status:
*^11: [1]
#: multiset
+2: [1,2]
-2: [1,2]
j1: [1]
opp1: [1]
*2: [1,2]


The following usable rules [14] were oriented: none



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

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

*1(1(x), y) → *1(x, y)
*1(0(x), y) → *1(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*1(1(x), y) → *1(x, y)
The remaining pairs can at least be oriented weakly.

*1(0(x), y) → *1(x, y)
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1, x2)
1(x1)  =  1(x1)
0(x1)  =  x1

Lexicographic path order with status [19].
Quasi-Precedence:
[*^12, 11]

Status:
*^12: [2,1]
11: [1]


The following usable rules [14] were oriented: none



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

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

*1(0(x), y) → *1(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


*1(0(x), y) → *1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1)
0(x1)  =  0(x1)

Lexicographic path order with status [19].
Quasi-Precedence:
01 > *^11

Status:
*^11: [1]
01: [1]


The following usable rules [14] were oriented: none



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

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

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(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.