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(#)), #)

Innermost 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), 0(y)) -> 0'(-(x, y))
-'(0(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(1(x), 1(y)) -> 0'(-(x, y))
-'(1(x), 1(y)) -> -'(x, y)
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), 0(y)) -> GE(x, y)
GE(1(x), 1(y)) -> GE(x, y)
GE(#, 0(x)) -> GE(#, x)
LOG(x) -> -'(log'(x), 1(#))
LOG(x) -> LOG'(x)
LOG'(1(x)) -> +'(log'(x), 1(#))
LOG'(1(x)) -> LOG'(x)
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)

Furthermore, R contains five SCCs.


   R
DPs
       →DP Problem 1
Usable Rules (Innermost)
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


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) -> #
-(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(#)), #)


Strategy:

innermost




As we are in the innermost case, we can delete all 21 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
           →DP Problem 6
Modular Removal of Rules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


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


Strategy:

innermost




We have the following set of usable rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
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
UsableRules
           →DP Problem 6
MRR
             ...
               →DP Problem 7
Modular Removal of Rules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




We have the following set of usable rules:

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


The result of this processor delivers one new DP problem.



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


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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
UsableRules
           →DP Problem 6
MRR
             ...
               →DP Problem 9
Modular Removal of Rules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


Dependency Pairs:

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


Rule:


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


Strategy:

innermost




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
UsableRules
           →DP Problem 6
MRR
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


Dependency Pair:

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


Rule:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
Usable Rules (Innermost)
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


Dependency Pairs:

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


Strategy:

innermost




As we are in the innermost case, we can delete all 22 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
           →DP Problem 11
Modular Removal of Rules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




We have the following set of usable rules:

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

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
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
Usable Rules (Innermost)
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


Dependency Pair:

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


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) -> #
-(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(#)), #)


Strategy:

innermost




As we are in the innermost case, we can delete all 29 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
           →DP Problem 12
Size-Change Principle
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




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
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
Usable Rules (Innermost)
       →DP Problem 5
UsableRules


Dependency Pairs:

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


Strategy:

innermost




As we are in the innermost case, we can delete all 29 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
           →DP Problem 13
Size-Change Principle
       →DP Problem 5
UsableRules


Dependency Pairs:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. GE(1(x), 1(y)) -> GE(x, y)
  2. GE(1(x), 0(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):
{1, 2, 3, 4} , {1, 2, 3, 4}
1>1
2>2
{1, 2, 3, 4} , {1, 2, 3, 4}
1>2
2>1

which lead(s) to this/these maximal multigraph(s):
{1, 2, 3, 4} , {1, 2, 3, 4}
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
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
Usable Rules (Innermost)


Dependency Pairs:

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


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) -> #
-(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(#)), #)


Strategy:

innermost




As we are in the innermost case, we can delete all 29 non-usable-rules.


   R
DPs
       →DP Problem 1
UsableRules
       →DP Problem 2
UsableRules
       →DP Problem 3
UsableRules
       →DP Problem 4
UsableRules
       →DP Problem 5
UsableRules
           →DP Problem 14
Size-Change Principle


Dependency Pairs:

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


Rule:

none


Strategy:

innermost




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

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

Innermost Termination of R successfully shown.
Duration:
0:02 minutes