Term Rewriting System R:
[x, y, z]
rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

REV(.(x, y)) -> ++'(rev(y), .(x, nil))
REV(.(x, y)) -> REV(y)
++'(.(x, y), z) -> ++'(y, z)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pair:

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


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cdr(x1))=  x1  
  POL(rev(x1))=  x1  
  POL(++'(x1, x2))=  x1  
  POL(null(x1))=  0  
  POL(false)=  0  
  POL(++(x1, x2))=  x1 + x2  
  POL(nil)=  0  
  POL(car(x1))=  1 + x1  
  POL(true)=  0  
  POL(.(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo


Dependency Pair:


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering


Dependency Pair:

REV(.(x, y)) -> REV(y)


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))





The following dependency pair can be strictly oriented:

REV(.(x, y)) -> REV(y)


Additionally, the following rules can be oriented:

rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(cdr(x1))=  x1  
  POL(rev(x1))=  x1  
  POL(REV(x1))=  x1  
  POL(null(x1))=  0  
  POL(false)=  0  
  POL(++(x1, x2))=  x1 + x2  
  POL(nil)=  0  
  POL(car(x1))=  1 + x1  
  POL(true)=  0  
  POL(.(x1, x2))=  1 + x1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph


Dependency Pair:


Rules:


rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))





Using the Dependency Graph resulted in no new DP problems.

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