Term Rewriting System R:
[x, y, z, l, l1, l2]
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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> 0'(+(+(x, y), 1(#)))
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
*'(0(x), y) -> 0'(*(x, y))
*'(0(x), y) -> *'(x, y)
*'(1(x), y) -> +'(0(*(x, y)), y)
*'(1(x), y) -> 0'(*(x, y))
*'(1(x), y) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)
*'(x, +(y, z)) -> +'(*(x, y), *(x, z))
*'(x, +(y, z)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
APP(cons(x, l1), l2) -> APP(l1, l2)
SUM(nil) -> 0'(#)
SUM(cons(x, l)) -> +'(x, sum(l))
SUM(cons(x, l)) -> SUM(l)
SUM(app(l1, l2)) -> +'(sum(l1), sum(l2))
SUM(app(l1, l2)) -> SUM(l1)
SUM(app(l1, l2)) -> SUM(l2)
PROD(cons(x, l)) -> *'(x, prod(l))
PROD(cons(x, l)) -> PROD(l)
PROD(app(l1, l2)) -> *'(prod(l1), prod(l2))
PROD(app(l1, l2)) -> PROD(l1)
PROD(app(l1, l2)) -> PROD(l2)

Furthermore, R contains five SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Modular Removal of Rules
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)


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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))





We have the following set of usable 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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  x1  
  POL(1(x1))=  x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {#, 0, 1, +, +'}
No Dependency Pairs can be deleted.
13 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 6
             ↳Modular Removal of Rules
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)


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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #





We have the following set of usable 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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  x1  
  POL(1(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 6
             ↳MRR
             ...
               →DP Problem 7
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(0(x), 0(y)) -> +'(x, y)


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))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #





We have the following set of usable 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))
+(+(x, y), z) -> +(x, +(y, z))
0(#) -> #
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  1 + x1  
  POL(1(x1))=  x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(0(x), 0(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 6
             ↳MRR
             ...
               →DP Problem 8
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))


Rules:


+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))





We have the following set of usable rules:

+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(#)=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {+, +'}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

+(x, #) -> x
+(#, x) -> x


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 6
             ↳MRR
             ...
               →DP Problem 9
                 ↳Modular Removal of Rules
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))


Rule:


+(+(x, y), z) -> +(x, +(y, z))





We have the following set of usable rules:

+(+(x, y), z) -> +(x, +(y, z))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

We have the following set D of usable symbols: {+, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(+(x, y), z) -> +'(y, z)

No Rules can be deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 6
             ↳MRR
             ...
               →DP Problem 10
                 ↳Size-Change Principle
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pair:

+'(+(x, y), z) -> +'(x, +(y, z))


Rule:


+(+(x, y), z) -> +(x, +(y, z))





We number the DPs as follows:
  1. +'(+(x, y), z) -> +'(x, +(y, z))
and get the following Size-Change Graph(s):
{1} , {1}
1>1

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial


We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳MRR
       →DP Problem 2
         ↳Size-Change Principle
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pair:

APP(cons(x, l1), l2) -> APP(l1, l2)


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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))





We number the DPs as follows:
  1. APP(cons(x, l1), l2) -> APP(l1, l2)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2=2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2=2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳MRR
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳Size-Change Principle
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳SCP


Dependency Pairs:

*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)
*'(*(x, y), z) -> *'(y, z)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)


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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))





We number the DPs as follows:
  1. *'(x, +(y, z)) -> *'(x, z)
  2. *'(x, +(y, z)) -> *'(x, y)
  3. *'(*(x, y), z) -> *'(y, z)
  4. *'(*(x, y), z) -> *'(x, *(y, z))
  5. *'(1(x), y) -> *'(x, y)
  6. *'(0(x), y) -> *'(x, y)
and get the following Size-Change Graph(s):
{6, 5, 4, 3, 2, 1} , {6, 5, 4, 3, 2, 1}
1=1
2>2
{6, 5, 4, 3, 2, 1} , {6, 5, 4, 3, 2, 1}
1>1
2=2
{6, 5, 4, 3, 2, 1} , {6, 5, 4, 3, 2, 1}
1>1

which lead(s) to this/these maximal multigraph(s):
{6, 5, 4, 3, 2, 1} , {6, 5, 4, 3, 2, 1}
1>1
2=2
{6, 5, 4, 3, 2, 1} , {6, 5, 4, 3, 2, 1}
1=1
2>2
{6, 5, 4, 3, 2, 1} , {6, 5, 4, 3, 2, 1}
1>1
{6, 5, 4, 3, 2, 1} , {6, 5, 4, 3, 2, 1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
0(x1) -> 0(x1)
1(x1) -> 1(x1)
+(x1, x2) -> +(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳MRR
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳Size-Change Principle
       →DP Problem 5
         ↳SCP


Dependency Pairs:

SUM(app(l1, l2)) -> SUM(l2)
SUM(app(l1, l2)) -> SUM(l1)
SUM(cons(x, l)) -> SUM(l)


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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))





We number the DPs as follows:
  1. SUM(app(l1, l2)) -> SUM(l2)
  2. SUM(app(l1, l2)) -> SUM(l1)
  3. SUM(cons(x, l)) -> SUM(l)
and get the following Size-Change Graph(s):
{3, 2, 1} , {3, 2, 1}
1>1

which lead(s) to this/these maximal multigraph(s):
{3, 2, 1} , {3, 2, 1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
app(x1, x2) -> app(x1, x2)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳MRR
       →DP Problem 2
         ↳SCP
       →DP Problem 3
         ↳SCP
       →DP Problem 4
         ↳SCP
       →DP Problem 5
         ↳Size-Change Principle


Dependency Pairs:

PROD(app(l1, l2)) -> PROD(l2)
PROD(app(l1, l2)) -> PROD(l1)
PROD(cons(x, l)) -> PROD(l)


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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
*(x, +(y, z)) -> +(*(x, y), *(x, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))





We number the DPs as follows:
  1. PROD(app(l1, l2)) -> PROD(l2)
  2. PROD(app(l1, l2)) -> PROD(l1)
  3. PROD(cons(x, l)) -> PROD(l)
and get the following Size-Change Graph(s):
{3, 2, 1} , {3, 2, 1}
1>1

which lead(s) to this/these maximal multigraph(s):
{3, 2, 1} , {3, 2, 1}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)
app(x1, x2) -> app(x1, x2)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:01 minutes