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:

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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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:

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 13 less nodes.

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

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

OPP(0(x)) → OPP(x)
OPP(j(x)) → OPP(x)
OPP(1(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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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.


OPP(0(x)) → OPP(x)
OPP(j(x)) → OPP(x)
OPP(1(x)) → OPP(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 4 + (4)x_1   
POL(j(x1)) = 4 + (3)x_1   
POL(0(x1)) = 4 + (4)x_1   
POL(OPP(x1)) = (4)x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ 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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

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

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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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.


+1(0(x), j(y)) → +1(x, y)
+1(j(x), 0(y)) → +1(x, y)
+1(1(x), j(y)) → +1(x, y)
+1(j(x), 1(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(x, y)
+1(0(x), 1(y)) → +1(x, y)
+1(1(x), 0(y)) → +1(x, y)
+1(j(x), j(y)) → +1(x, y)
+1(1(x), 1(y)) → +1(+(x, y), 1(#))
+1(j(x), j(y)) → +1(+(x, y), j(#))
The remaining pairs can at least be oriented weakly.

+1(+(x, y), z) → +1(y, z)
+1(0(x), 0(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 2 + x_1   
POL(j(x1)) = 2 + x_1   
POL(0(x1)) = x_1   
POL(+1(x1, x2)) = (4)x_1 + (4)x_2   
POL(+(x1, x2)) = x_1 + x_2   
POL(#) = 0   
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:

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



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

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

+1(0(x), 0(y)) → +1(x, y)
+1(+(x, y), z) → +1(y, z)
+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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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.


+1(0(x), 0(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.

+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 2 + (13/4)x_1   
POL(j(x1)) = 1/4   
POL(0(x1)) = 1/4 + x_1   
POL(+1(x1, x2)) = (4)x_1   
POL(+(x1, x2)) = (2)x_1 + (11/4)x_2   
POL(#) = 0   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



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

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

+1(+(x, y), z) → +1(y, z)
+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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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.


+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 13/4   
POL(j(x1)) = 3   
POL(0(x1)) = 4   
POL(+1(x1, x2)) = (1/4)x_1   
POL(+(x1, x2)) = 4 + (2)x_1 + (9/4)x_2   
POL(#) = 1/4   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
          ↳ 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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

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

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


*1(+(x, y), z) → *1(x, z)
*1(j(x), y) → *1(x, y)
*1(+(x, y), z) → *1(y, z)
*1(0(x), y) → *1(x, y)
*1(1(x), y) → *1(x, y)
The remaining pairs can at least be oriented weakly.

*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
*1(*(x, y), z) → *1(y, z)
*1(*(x, y), z) → *1(x, *(y, z))
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 3 + (3/2)x_1   
POL(j(x1)) = 13/4 + x_1   
POL(*1(x1, x2)) = (2)x_1   
POL(opp(x1)) = 3/4 + (4)x_1   
POL(-(x1, x2)) = 1/2 + (7/2)x_2   
POL(*(x1, x2)) = x_1 + (2)x_2   
POL(0(x1)) = 11/4 + x_1   
POL(+(x1, x2)) = 3 + x_1 + (2)x_2   
POL(#) = 4   
The value of delta used in the strict ordering is 11/2.
The following usable rules [17] were oriented: none



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

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

*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
*1(*(x, y), z) → *1(y, z)
*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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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.


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

*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
Used ordering: Polynomial interpretation [25,35]:

POL(1(x1)) = 4 + (7/4)x_1   
POL(j(x1)) = 7/4 + (7/4)x_1   
POL(*1(x1, x2)) = x_1   
POL(opp(x1)) = 5/4 + (4)x_1   
POL(-(x1, x2)) = 3 + (2)x_2   
POL(*(x1, x2)) = 1/4 + (4)x_1 + (2)x_2   
POL(0(x1)) = x_1   
POL(+(x1, x2)) = 4   
POL(#) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none



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

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

*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, 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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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.


*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(*1(x1, x2)) = (4)x_2   
POL(+(x1, x2)) = 1 + (4)x_1 + (4)x_2   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, 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.