Term Rewriting System R:
[l, x, k, a, b, c]
f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(cons(x, k), l) -> G(k, l, cons(x, k))
G(a, b, c) -> F(a, cons(b, c))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Instantiation Transformation


Dependency Pairs:

G(a, b, c) -> F(a, cons(b, c))
F(cons(x, k), l) -> G(k, l, cons(x, k))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))





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

F(cons(x, k), l) -> G(k, l, cons(x, k))
one new Dependency Pair is created:

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 2
             ↳Instantiation Transformation


Dependency Pairs:

F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))
G(a, b, c) -> F(a, cons(b, c))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))





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

G(a, b, c) -> F(a, cons(b, c))
one new Dependency Pair is created:

G(a', cons(b'''', c''''), cons(x''', k'''')) -> F(a', cons(cons(b'''', c''''), cons(x''', k'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Inst
           →DP Problem 2
             ↳Inst
             ...
               →DP Problem 3
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

G(a', cons(b'''', c''''), cons(x''', k'''')) -> F(a', cons(cons(b'''', c''''), cons(x''', k'''')))
F(cons(x', k'), cons(b'', c'')) -> G(k', cons(b'', c''), cons(x', k'))


Rules:


f(empty, l) -> l
f(cons(x, k), l) -> g(k, l, cons(x, k))
g(a, b, c) -> f(a, cons(b, c))




Termination of R could not be shown.
Duration:
0:05 minutes