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