Term Rewriting System R:
[x, y, z]
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) -> #
-(x, #) -> x
-(0(x), 0(y)) -> 0(-(x, y))
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) -> 1(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
not(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 0(y)) -> ge(x, y)
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)

Termination of R to be shown. 



   R
     ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

log(x) -> -(log'(x), 1(#))

where the Polynomial interpretation:
  POL(#)=  0  
  POL(if(x1, x2, x3))=  x1 + x2 + x3  
  POL(0(x1))=  2·x1  
  POL(false)=  0  
  POL(log'(x1))=  x1  
  POL(1(x1))=  2·x1  
  POL(log(x1))=  1 + x1  
  POL(true)=  0  
  POL(ge(x1, x2))=  x1 + x2  
  POL(-(x1, x2))=  x1 + x2  
  POL(not(x1))=  x1  
  POL(+(x1, x2))=  x1 + x2  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

GE(0(x), 0(y)) -> GE(x, y)
GE(0(x), 1(y)) -> NOT(ge(y, x))
GE(0(x), 1(y)) -> GE(y, x)
GE(1(x), 1(y)) -> GE(x, y)
GE(#, 0(x)) -> GE(#, x)
GE(1(x), 0(y)) -> GE(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 1(y)) -> -'(x, y)
-'(0(x), 0(y)) -> 0'(-(x, y))
-'(0(x), 0(y)) -> -'(x, y)
-'(1(x), 1(y)) -> 0'(-(x, y))
-'(1(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(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)
LOG'(0(x)) -> IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
LOG'(0(x)) -> GE(x, 1(#))
LOG'(0(x)) -> +'(log'(x), 1(#))
LOG'(0(x)) -> LOG'(x)
LOG'(1(x)) -> +'(log'(x), 1(#))
LOG'(1(x)) -> LOG'(x)

Furthermore, R contains five SCCs. 


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Size-Change Principle
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
           →DP Problem 4
             ↳SCP
           →DP Problem 5
             ↳SCP


Dependency Pair:

GE(#, 0(x)) -> GE(#, x)


Rules:


ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(1(x), 0(y)) -> ge(x, y)
ge(#, 1(x)) -> false
0(#) -> #
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
-(1(x), 0(y)) -> 1(-(x, y))
-(x, #) -> x
-(#, x) -> #
not(true) -> false
not(false) -> true
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
log'(1(x)) -> +(log'(x), 1(#))
log'(#) -> #
if(false, x, y) -> y
if(true, x, y) -> x





We number the DPs as follows:
  1. GE(#, 0(x)) -> GE(#, x)
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:
0(x1) -> 0(x1)

We obtain no new DP problems. 


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳Modular Removal of Rules
           →DP Problem 3
             ↳MRR
           →DP Problem 4
             ↳SCP
           →DP Problem 5
             ↳SCP


Dependency Pairs:

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


Rules:


ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(1(x), 0(y)) -> ge(x, y)
ge(#, 1(x)) -> false
0(#) -> #
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
-(1(x), 0(y)) -> 1(-(x, y))
-(x, #) -> x
-(#, x) -> #
not(true) -> false
not(false) -> true
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
log'(1(x)) -> +(log'(x), 1(#))
log'(#) -> #
if(false, x, y) -> y
if(true, x, y) -> x





We have the following set of usable rules:

-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
-(1(x), 0(y)) -> 1(-(x, y))
-(x, #) -> x
-(#, x) -> #
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(-'(x1, x2))=  1 + x1 + x2  
  POL(0(x1))=  x1  
  POL(1(x1))=  x1  
  POL(-(x1, x2))=  x1 + x2  

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

The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
             ...
               →DP Problem 6
                 ↳Modular Removal of Rules
           →DP Problem 3
             ↳MRR
           →DP Problem 4
             ↳SCP
           →DP Problem 5
             ↳SCP


Dependency Pairs:

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


Rules:


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





We have the following set of usable rules:

-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
-(1(x), 0(y)) -> 1(-(x, y))
-(x, #) -> x
-(#, x) -> #
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(-'(x1, x2))=  1 + x1 + x2  
  POL(0(x1))=  1 + x1  
  POL(1(x1))=  1 + x1  
  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:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved. 



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳Modular Removal of Rules
           →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(#))
+'(0(x), 0(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)


Rules:


ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(1(x), 0(y)) -> ge(x, y)
ge(#, 1(x)) -> false
0(#) -> #
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
-(1(x), 0(y)) -> 1(-(x, y))
-(x, #) -> x
-(#, x) -> #
not(true) -> false
not(false) -> true
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
log'(1(x)) -> +(log'(x), 1(#))
log'(#) -> #
if(false, x, y) -> y
if(true, x, y) -> x





We have the following set of usable rules:

+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
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.
20 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
             ...
               →DP Problem 7
                 ↳Modular Removal of Rules
           →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(#))
+'(0(x), 0(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)


Rules:


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





We have the following set of usable rules:

+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
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
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
             ...
               →DP Problem 8
                 ↳Modular Removal of Rules
           →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:


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





We have the following set of usable rules:

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


The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
             ...
               →DP Problem 9
                 ↳Modular Removal of Rules
           →DP Problem 4
             ↳SCP
           →DP Problem 5
             ↳SCP


Dependency Pairs:

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


Rules:


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





We have the following set of usable rules:

+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
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
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
             ...
               →DP Problem 10
                 ↳Modular Removal of Rules
           →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
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
             ...
               →DP Problem 11
                 ↳Size-Change Principle
           →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
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
           →DP Problem 4
             ↳Size-Change Principle
           →DP Problem 5
             ↳SCP


Dependency Pairs:

GE(1(x), 0(y)) -> GE(x, y)
GE(1(x), 1(y)) -> GE(x, y)
GE(0(x), 1(y)) -> GE(y, x)
GE(0(x), 0(y)) -> GE(x, y)


Rules:


ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(1(x), 0(y)) -> ge(x, y)
ge(#, 1(x)) -> false
0(#) -> #
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
-(1(x), 0(y)) -> 1(-(x, y))
-(x, #) -> x
-(#, x) -> #
not(true) -> false
not(false) -> true
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
log'(1(x)) -> +(log'(x), 1(#))
log'(#) -> #
if(false, x, y) -> y
if(true, x, y) -> x





We number the DPs as follows:
  1. GE(1(x), 0(y)) -> GE(x, y)
  2. GE(1(x), 1(y)) -> GE(x, y)
  3. GE(0(x), 1(y)) -> GE(y, x)
  4. GE(0(x), 0(y)) -> GE(x, y)
and get the following Size-Change Graph(s):
{4, 3, 2, 1} , {4, 3, 2, 1}
1>1
2>2
{4, 3, 2, 1} , {4, 3, 2, 1}
1>2
2>1

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

We obtain no new DP problems. 


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳SCP
           →DP Problem 2
             ↳MRR
           →DP Problem 3
             ↳MRR
           →DP Problem 4
             ↳SCP
           →DP Problem 5
             ↳Size-Change Principle


Dependency Pairs:

LOG'(1(x)) -> LOG'(x)
LOG'(0(x)) -> LOG'(x)


Rules:


ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(1(x), 0(y)) -> ge(x, y)
ge(#, 1(x)) -> false
0(#) -> #
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(0(x), 0(y)) -> 0(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
-(1(x), 0(y)) -> 1(-(x, y))
-(x, #) -> x
-(#, x) -> #
not(true) -> false
not(false) -> true
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), 0(y)) -> 0(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
+(x, #) -> x
+(#, x) -> x
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)
log'(1(x)) -> +(log'(x), 1(#))
log'(#) -> #
if(false, x, y) -> y
if(true, x, y) -> x





We number the DPs as follows:
  1. LOG'(1(x)) -> LOG'(x)
  2. LOG'(0(x)) -> LOG'(x)
and get the following Size-Change Graph(s):
{2, 1} , {2, 1}
1>1

which lead(s) to this/these maximal multigraph(s):
{2, 1} , {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:
0(x1) -> 0(x1)
1(x1) -> 1(x1)

We obtain no new DP problems.

Termination of R successfully shown.
Duration:
0:07 minutes