(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(1(x)) → OPP(x)
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)  =  x1
1(x1)  =  1(x1)
0(x1)  =  0(x1)
j(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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))
*(+(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(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)  =  x1
j(x1)  =  j(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) 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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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.

(13) 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.

(14) 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)
*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)  =  0(x1)
j(x1)  =  x1
*(x1, x2)  =  *(x1, x2)
+(x1, x2)  =  +(x1, x2)
#  =  #
-(x1, x2)  =  -(x1, x2)
opp(x1)  =  opp

Lexicographic Path Order [LPO].
Precedence:
[*2, #] > *^11 > [01, +2, opp]
[*2, #] > -2 > [01, +2, opp]


The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

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

*1(1(x), y) → *1(x, y)
*1(j(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.

(16) 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(x2)
1(x1)  =  1(x1)
j(x1)  =  j(x1)
+(x1, x2)  =  +(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


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(j(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.

(18) 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)
j(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(19) Obligation:

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

*1(j(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.

(20) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x1
j(x1)  =  j(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(21) 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.

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE