Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 QDPSizeChangeProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 UsableRulesProof (⇔)
    ↳12 QDP
    ↳13 MRRProof (⇔)
    ↳14 QDP
    ↳15 QDPSizeChangeProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPSizeChangeProof (⇔)
    ↳19 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))

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)

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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 11 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))

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

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) 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)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • OPP(1(x)) → OPP(x)
    The graph contains the following edges 1 > 1

  • OPP(0(x)) → OPP(x)
    The graph contains the following edges 1 > 1

  • OPP(j(x)) → OPP(x)
    The graph contains the following edges 1 > 1

(9) TRUE

(10) 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))

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

(11) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(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:

+(#, 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))
0(#) → #

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

(13) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

+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)) → +1(x, y)
+1(j(x), 1(y)) → +1(x, y)

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

POL(#) = 0   
POL(+(x1, x2)) = x1 + x2   
POL(+1(x1, x2)) = x1 + x2   
POL(0(x1)) = x1   
POL(1(x1)) = 1 + x1   
POL(j(x1)) = 1 + x1   

(14) Obligation:

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

+1(0(x), 0(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:

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

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

(15) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • +1(0(x), 0(y)) → +1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

  • +1(+(x, y), z) → +1(x, +(y, z))
    The graph contains the following edges 1 > 1

  • +1(+(x, y), z) → +1(y, z)
    The graph contains the following edges 1 > 1, 2 >= 2

(16) TRUE

(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(j(x), 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))

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

(18) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • *1(1(x), y) → *1(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

  • *1(0(x), y) → *1(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

  • *1(j(x), y) → *1(x, y)
    The graph contains the following edges 1 > 1, 2 >= 2

  • *1(*(x, y), z) → *1(x, *(y, z))
    The graph contains the following edges 1 > 1

  • *1(*(x, y), z) → *1(y, z)
    The graph contains the following edges 1 > 1, 2 >= 2

(19) TRUE