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
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
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


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(#)), #)





The following dependency pairs can be strictly oriented:

+'(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)


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

+(#, 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(#) -> #


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  x1  
  POL(1(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


Dependency Pairs:

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





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  1 + x1  
  POL(1(x1))=  0  
  POL(+(x1, x2))=  x1 + x2  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polo
             ...
               →DP Problem 7
Polynomial Ordering
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


Dependency Pairs:

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


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(#)), #)





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  0  
  POL(1(x1))=  0  
  POL(+(x1, x2))=  1 + x1 + x2  
  POL(+'(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
           →DP Problem 6
Polo
             ...
               →DP Problem 8
Dependency Graph
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


Dependency Pair:


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(#)), #)





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


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(#)), #)





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  1 + x1  
  POL(-'(x1, x2))=  x2  
  POL(1(x1))=  x1  
  POL(-(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 9
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


Dependency Pairs:

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


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(#)), #)





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  0  
  POL(-'(x1, x2))=  x2  
  POL(1(x1))=  1 + x1  
  POL(-(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 9
Polo
             ...
               →DP Problem 10
Polynomial Ordering
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


Dependency Pair:

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


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(#)), #)





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

-(#, 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))
0(#) -> #


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  1 + x1  
  POL(-'(x1, x2))=  1 + x1  
  POL(1(x1))=  1 + x1  
  POL(-(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
           →DP Problem 9
Polo
             ...
               →DP Problem 11
Dependency Graph
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo


Dependency Pair:


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(#)), #)





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polynomial Ordering
       →DP Problem 4
Polo
       →DP Problem 5
Polo


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(#)), #)





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(#)=  0  
  POL(0(x1))=  1 + x1  
  POL(GE(x1, x2))=  x2  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
           →DP Problem 12
Dependency Graph
       →DP Problem 4
Polo
       →DP Problem 5
Polo


Dependency Pair:


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(#)), #)





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polynomial Ordering
       →DP Problem 5
Polo


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(#)), #)





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0(x1))=  x1  
  POL(GE(x1, x2))=  1 + x1 + x2  
  POL(1(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 13
Polynomial Ordering
       →DP Problem 5
Polo


Dependency Pair:

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(#)), #)





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0(x1))=  1 + x1  
  POL(GE(x1, x2))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
           →DP Problem 13
Polo
             ...
               →DP Problem 14
Dependency Graph
       →DP Problem 5
Polo


Dependency Pair:


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(#)), #)





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polynomial Ordering


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(#)), #)





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LOG'(x1))=  x1  
  POL(0(x1))=  1 + x1  
  POL(1(x1))=  x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
           →DP Problem 15
Polynomial Ordering


Dependency Pair:

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(#)), #)





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LOG'(x1))=  x1  
  POL(1(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Polo
       →DP Problem 3
Polo
       →DP Problem 4
Polo
       →DP Problem 5
Polo
           →DP Problem 15
Polo
             ...
               →DP Problem 16
Dependency Graph


Dependency Pair:


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(#)), #)





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes