Term Rewriting System R:
[y, x, z]
f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(.(nil, y)) -> F(y)
F(.(.(x, y), z)) -> F(.(x, .(y, z)))
G(.(x, nil)) -> G(x)
G(.(x, .(y, z))) -> G(.(.(x, y), z))

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))
F(.(nil, y)) -> F(y)


Rules:


f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))





The following dependency pair can be strictly oriented:

F(.(nil, y)) -> F(y)


The following rules can be oriented:

f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(g(x1))=  x1  
  POL(nil)=  1  
  POL(.(x1, x2))=  x1 + x2  
  POL(F(x1))=  1 + x1  
  POL(f(x1))=  x1  

resulting in one new DP problem.
Used Argument Filtering System:
F(x1) -> F(x1)
.(x1, x2) -> .(x1, x2)
f(x1) -> f(x1)
g(x1) -> g(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

F(.(.(x, y), z)) -> F(.(x, .(y, z)))


Rules:


f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(.(.(x, y), z)) -> F(.(x, .(y, z)))
one new Dependency Pair is created:

F(.(.(x'', y0), .(y'', z''))) -> F(.(x'', .(y0, .(y'', z''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 4
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

F(.(.(x'', y0), .(y'', z''))) -> F(.(x'', .(y0, .(y'', z''))))


Rules:


f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(.(.(x'', y0), .(y'', z''))) -> F(.(x'', .(y0, .(y'', z''))))
one new Dependency Pair is created:

F(.(.(x'''', y0''), .(y''0, .(y'''', z'''')))) -> F(.(x'''', .(y0'', .(y''0, .(y'''', z'''')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 5
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

F(.(.(x'''', y0''), .(y''0, .(y'''', z'''')))) -> F(.(x'''', .(y0'', .(y''0, .(y'''', z'''')))))


Rules:


f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(.(.(x'''', y0''), .(y''0, .(y'''', z'''')))) -> F(.(x'''', .(y0'', .(y''0, .(y'''', z'''')))))
one new Dependency Pair is created:

F(.(.(x'''''', y0''''), .(y''0'', .(y''''0, .(y'''''', z''''''))))) -> F(.(x'''''', .(y0'''', .(y''0'', .(y''''0, .(y'''''', z''''''))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 6
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

F(.(.(x'''''', y0''''), .(y''0'', .(y''''0, .(y'''''', z''''''))))) -> F(.(x'''''', .(y0'''', .(y''0'', .(y''''0, .(y'''''', z''''''))))))


Rules:


f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(.(.(x'''''', y0''''), .(y''0'', .(y''''0, .(y'''''', z''''''))))) -> F(.(x'''''', .(y0'''', .(y''0'', .(y''''0, .(y'''''', z''''''))))))
one new Dependency Pair is created:

F(.(.(x'''''''', y0''''''), .(y''0'''', .(y''''0'', .(y'''''''', .(y''''''''', z'''''''')))))) -> F(.(x'''''''', .(y0'''''', .(y''0'''', .(y''''0'', .(y'''''''', .(y''''''''', z'''''''')))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 7
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Remaining


Dependency Pair:

F(.(.(x'''''''', y0''''''), .(y''0'''', .(y''''0'', .(y'''''''', .(y''''''''', z'''''''')))))) -> F(.(x'''''''', .(y0'''''', .(y''0'''', .(y''''0'', .(y'''''''', .(y''''''''', z'''''''')))))))


Rules:


f(nil) -> nil
f(.(nil, y)) -> .(nil, f(y))
f(.(.(x, y), z)) -> f(.(x, .(y, z)))
g(nil) -> nil
g(.(x, nil)) -> .(g(x), nil)
g(.(x, .(y, z))) -> g(.(.(x, y), z))





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

F(.(.(x'''''''', y0''''''), .(y''0'''', .(y''''0'', .(y'''''''', .(y''''''''', z'''''''')))))) -> F(.(x'''''''', .(y0'''''', .(y''0'''', .(y''''0'', .(y'''''''', .(y''''''''', z'''''''')))))))
one new Dependency Pair is created:

F(.(.(x'''''''''', y0''''''''), .(y''0'''''', .(y''''0'''', .(y''''''''0, .(y'''''''''0, .(y'''''''''''', z''''''''''))))))) -> F(.(x'''''''''', .(y0'''''''', .(y''0'''''', .(y''''0'''', .(y''''''''0, .(y'''''''''0, .(y'''''''''''', z''''''''''))))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Remaining Obligation(s)




The following remains to be proven:

Termination of R could not be shown.
Duration:
0:00 minutes