Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 PisEmptyProof (⇔)
    ↳13 TRUE
  ↳14 QDP
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 PisEmptyProof (⇔)
    ↳25 TRUE

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 13 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

OPP(1(x)) → OPP(x)
OPP(0(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))
*(+(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.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


OPP(j(x)) → OPP(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
OPP(x1)  =  OPP(x1)
1(x1)  =  x1
0(x1)  =  x1
j(x1)  =  j(x1)

Lexicographic path order with status [LPO].
Precedence:
j1 > OPP1

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

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

OPP(1(x)) → OPP(x)
OPP(0(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.

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


OPP(0(x)) → OPP(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
OPP(x1)  =  OPP(x1)
1(x1)  =  x1
0(x1)  =  0(x1)

Lexicographic path order with status [LPO].
Precedence:
01 > OPP1

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

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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.

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


OPP(1(x)) → OPP(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
OPP(x1)  =  x1
1(x1)  =  1(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
11: [1]

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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.

(12) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(13) TRUE

(14) Obligation:

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

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

(15) Obligation:

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


*1(j(x), y) → *1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
*1(*(x, y), z) → *1(y, z)
*1(+(x, y), z) → *1(x, z)
*1(+(x, y), z) → *1(y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1)
1(x1)  =  x1
0(x1)  =  x1
j(x1)  =  j(x1)
*(x1, x2)  =  *(x1, x2)
+(x1, x2)  =  +(x1, x2)
#  =  #
-(x1, x2)  =  -(x1)
opp(x1)  =  opp(x1)

Lexicographic path order with status [LPO].
Precedence:
*2 > +2
-1 > +2
opp1 > j1
opp1 > #

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

The following usable rules [FROCOS05] were oriented: none

(17) Obligation:

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)
*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.

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
01: [1]

The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

*1(1(x), y) → *1(x, y)
*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.

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1, x2)
1(x1)  =  x1
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Precedence:
trivial

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

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

*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.

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
11: [1]

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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.

(24) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(25) TRUE