Term Rewriting System R:
[x, a, k, y]
f(x, empty) -> x
f(empty, cons(a, k)) -> f(cons(a, k), k)
f(cons(a, k), y) -> f(y, k)

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(empty, cons(a, k)) -> F(cons(a, k), k)
F(cons(a, k), y) -> F(y, k)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation


Dependency Pairs:

F(cons(a, k), y) -> F(y, k)
F(empty, cons(a, k)) -> F(cons(a, k), k)


Rules:


f(x, empty) -> x
f(empty, cons(a, k)) -> f(cons(a, k), k)
f(cons(a, k), y) -> f(y, k)


Strategy:

innermost




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

F(cons(a, k), y) -> F(y, k)
two new Dependency Pairs are created:

F(cons(a, k''), cons(a'', k''')) -> F(cons(a'', k'''), k'')
F(cons(a, cons(a'', k'')), empty) -> F(empty, cons(a'', k''))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

F(cons(a, cons(a'', k'')), empty) -> F(empty, cons(a'', k''))
F(cons(a, k''), cons(a'', k''')) -> F(cons(a'', k'''), k'')
F(empty, cons(a, k)) -> F(cons(a, k), k)


Rules:


f(x, empty) -> x
f(empty, cons(a, k)) -> f(cons(a, k), k)
f(cons(a, k), y) -> f(y, k)


Strategy:

innermost




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

F(empty, cons(a, k)) -> F(cons(a, k), k)
one new Dependency Pair is created:

F(empty, cons(a'', cons(a'''', k''''''))) -> F(cons(a'', cons(a'''', k'''''')), cons(a'''', k''''''))

The transformation is resulting in one new DP problem: 



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




The following remains to be proven:
Dependency Pairs:

F(cons(a, k''), cons(a'', k''')) -> F(cons(a'', k'''), k'')
F(empty, cons(a'', cons(a'''', k''''''))) -> F(cons(a'', cons(a'''', k'''''')), cons(a'''', k''''''))
F(cons(a, cons(a'', k'')), empty) -> F(empty, cons(a'', k''))


Rules:


f(x, empty) -> x
f(empty, cons(a, k)) -> f(cons(a, k), k)
f(cons(a, k), y) -> f(y, k)


Strategy:

innermost



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