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))

Innermost 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
         ↳FwdInst


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))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst


Dependency Pair:


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))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Forward Instantiation Transformation


Dependency Pairs:

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


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))


Strategy:

innermost




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

G(.(x, nil)) -> G(x)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳FwdInst
           →DP Problem 4
             ↳Forward Instantiation Transformation


Dependency Pairs:

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


Strategy:

innermost




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

G(.(x, .(y, z))) -> G(.(.(x, y), z))
three new Dependency Pairs are created:

G(.(x'', .(y0, .(y'', z'')))) -> G(.(.(x'', y0), .(y'', z'')))
G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))
G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳FwdInst
           →DP Problem 4
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

G(.(x', .(.(y'''', z''''), nil))) -> G(.(.(x', .(y'''', z'''')), nil))
G(.(.(x'', nil), nil)) -> G(.(x'', nil))
G(.(x', .(nil, nil))) -> G(.(.(x', nil), nil))
G(.(x'', .(y0, .(y'', z'')))) -> G(.(.(x'', y0), .(y'', z'')))
G(.(.(x'', .(y'', z'')), nil)) -> G(.(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))


Strategy:

innermost



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