Term Rewriting System R:
[x, y]
fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

FAC(0) -> 1'
FAC(s(x)) -> *'(s(x), fac(x))
FAC(s(x)) -> FAC(x)
FLOOP(s(x), y) -> FLOOP(x, *(s(x), y))
FLOOP(s(x), y) -> *'(s(x), y)
*'(x, s(y)) -> +'(*(x, y), x)
*'(x, s(y)) -> *'(x, y)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains four SCCs.


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


Dependency Pair:

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


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





The following dependency pair can be strictly oriented:

+'(x, s(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(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem.



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


Dependency Pair:


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





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


Dependency Pair:

*'(x, s(y)) -> *'(x, y)


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





The following dependency pair can be strictly oriented:

*'(x, s(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(*'(x1, x2))=  x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



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


Dependency Pair:


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





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


Dependency Pair:

FAC(s(x)) -> FAC(x)


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





The following dependency pair can be strictly oriented:

FAC(s(x)) -> FAC(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(FAC(x1))=  x1  
  POL(s(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 7
Dependency Graph
       →DP Problem 4
Polo


Dependency Pair:


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





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


Dependency Pair:

FLOOP(s(x), y) -> FLOOP(x, *(s(x), y))


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





The following dependency pair can be strictly oriented:

FLOOP(s(x), y) -> FLOOP(x, *(s(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)=  0  
  POL(*(x1, x2))=  0  
  POL(s(x1))=  1 + x1  
  POL(FLOOP(x1, x2))=  x1  
  POL(+(x1, x2))=  0  

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 8
Dependency Graph


Dependency Pair:


Rules:


fac(0) -> 1
fac(s(x)) -> *(s(x), fac(x))
fac(0) -> s(0)
floop(0, y) -> y
floop(s(x), y) -> floop(x, *(s(x), y))
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
1 -> s(0)





Using the Dependency Graph resulted in no new DP problems.

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