Termination Proof using AProVE

Term Rewriting System R:
[Z, X, Y, X1, X2]
a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_FST}(s(X), cons(Y, Z)) -> MARK(Y)
A_{_FROM}(X) -> MARK(X)
A_{_ADD}(0, X) -> MARK(X)
MARK(fst(X1, X2)) -> A_{_FST}(mark(X1), mark(X2))
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> MARK(X2)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(from(X)) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))
MARK(add(X1, X2)) -> MARK(X1)
MARK(add(X1, X2)) -> MARK(X2)
MARK(len(X)) -> A_{_LEN}(mark(X))
MARK(len(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(len(X)) -> MARK(X)
MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(fst(X1, X2)) -> A_{_FST}(mark(X1), mark(X2))
A_{_FST}(s(X), cons(Y, Z)) -> MARK(Y)

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

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

MARK(fst(X1, X2)) -> A_{_FST}(mark(X1), mark(X2))

16 new Dependency Pairs are created:

MARK(fst(fst(X1'', X2''), X2)) -> A_{_FST}(a_{_fst}(mark(X1''), mark(X2'')), mark(X2))
MARK(fst(from(X'), X2)) -> A_{_FST}(a_{_from}(mark(X')), mark(X2))
MARK(fst(add(X1'', X2''), X2)) -> A_{_FST}(a_{_add}(mark(X1''), mark(X2'')), mark(X2))
MARK(fst(len(X'), X2)) -> A_{_FST}(a_{_len}(mark(X')), mark(X2))
MARK(fst(0, X2)) -> A_{_FST}(0, mark(X2))
MARK(fst(s(X'), X2)) -> A_{_FST}(s(X'), mark(X2))
MARK(fst(nil, X2)) -> A_{_FST}(nil, mark(X2))
MARK(fst(cons(X1'', X2''), X2)) -> A_{_FST}(cons(mark(X1''), X2''), mark(X2))
MARK(fst(X1, fst(X1'', X2''))) -> A_{_FST}(mark(X1), a_{_fst}(mark(X1''), mark(X2'')))
MARK(fst(X1, from(X'))) -> A_{_FST}(mark(X1), a_{_from}(mark(X')))
MARK(fst(X1, add(X1'', X2''))) -> A_{_FST}(mark(X1), a_{_add}(mark(X1''), mark(X2'')))
MARK(fst(X1, len(X'))) -> A_{_FST}(mark(X1), a_{_len}(mark(X')))
MARK(fst(X1, 0)) -> A_{_FST}(mark(X1), 0)
MARK(fst(X1, s(X'))) -> A_{_FST}(mark(X1), s(X'))
MARK(fst(X1, nil)) -> A_{_FST}(mark(X1), nil)
MARK(fst(X1, cons(X1'', X2''))) -> A_{_FST}(mark(X1), cons(mark(X1''), X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

MARK(fst(X1, cons(X1'', X2''))) -> A_{_FST}(mark(X1), cons(mark(X1''), X2''))
MARK(fst(X1, len(X'))) -> A_{_FST}(mark(X1), a_{_len}(mark(X')))
MARK(fst(X1, add(X1'', X2''))) -> A_{_FST}(mark(X1), a_{_add}(mark(X1''), mark(X2'')))
MARK(fst(X1, from(X'))) -> A_{_FST}(mark(X1), a_{_from}(mark(X')))
MARK(fst(X1, fst(X1'', X2''))) -> A_{_FST}(mark(X1), a_{_fst}(mark(X1''), mark(X2'')))
MARK(fst(s(X'), X2)) -> A_{_FST}(s(X'), mark(X2))
MARK(fst(len(X'), X2)) -> A_{_FST}(a_{_len}(mark(X')), mark(X2))
MARK(fst(add(X1'', X2''), X2)) -> A_{_FST}(a_{_add}(mark(X1''), mark(X2'')), mark(X2))
MARK(fst(from(X'), X2)) -> A_{_FST}(a_{_from}(mark(X')), mark(X2))
A_{_FST}(s(X), cons(Y, Z)) -> MARK(Y)
MARK(fst(fst(X1'', X2''), X2)) -> A_{_FST}(a_{_fst}(mark(X1''), mark(X2'')), mark(X2))
MARK(len(X)) -> MARK(X)
MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
A_{_ADD}(0, X) -> MARK(X)
MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

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

MARK(add(X1, X2)) -> A_{_ADD}(mark(X1), mark(X2))

16 new Dependency Pairs are created:

MARK(add(fst(X1'', X2''), X2)) -> A_{_ADD}(a_{_fst}(mark(X1''), mark(X2'')), mark(X2))
MARK(add(from(X'), X2)) -> A_{_ADD}(a_{_from}(mark(X')), mark(X2))
MARK(add(add(X1'', X2''), X2)) -> A_{_ADD}(a_{_add}(mark(X1''), mark(X2'')), mark(X2))
MARK(add(len(X'), X2)) -> A_{_ADD}(a_{_len}(mark(X')), mark(X2))
MARK(add(0, X2)) -> A_{_ADD}(0, mark(X2))
MARK(add(s(X'), X2)) -> A_{_ADD}(s(X'), mark(X2))
MARK(add(nil, X2)) -> A_{_ADD}(nil, mark(X2))
MARK(add(cons(X1'', X2''), X2)) -> A_{_ADD}(cons(mark(X1''), X2''), mark(X2))
MARK(add(X1, fst(X1'', X2''))) -> A_{_ADD}(mark(X1), a_{_fst}(mark(X1''), mark(X2'')))
MARK(add(X1, from(X'))) -> A_{_ADD}(mark(X1), a_{_from}(mark(X')))
MARK(add(X1, add(X1'', X2''))) -> A_{_ADD}(mark(X1), a_{_add}(mark(X1''), mark(X2'')))
MARK(add(X1, len(X'))) -> A_{_ADD}(mark(X1), a_{_len}(mark(X')))
MARK(add(X1, 0)) -> A_{_ADD}(mark(X1), 0)
MARK(add(X1, s(X'))) -> A_{_ADD}(mark(X1), s(X'))
MARK(add(X1, nil)) -> A_{_ADD}(mark(X1), nil)
MARK(add(X1, cons(X1'', X2''))) -> A_{_ADD}(mark(X1), cons(mark(X1''), X2''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(add(X1, cons(X1'', X2''))) -> A_{_ADD}(mark(X1), cons(mark(X1''), X2''))
MARK(add(X1, nil)) -> A_{_ADD}(mark(X1), nil)
MARK(add(X1, s(X'))) -> A_{_ADD}(mark(X1), s(X'))
MARK(add(X1, 0)) -> A_{_ADD}(mark(X1), 0)
MARK(add(X1, len(X'))) -> A_{_ADD}(mark(X1), a_{_len}(mark(X')))
MARK(add(X1, add(X1'', X2''))) -> A_{_ADD}(mark(X1), a_{_add}(mark(X1''), mark(X2'')))
MARK(add(X1, from(X'))) -> A_{_ADD}(mark(X1), a_{_from}(mark(X')))
MARK(add(X1, fst(X1'', X2''))) -> A_{_ADD}(mark(X1), a_{_fst}(mark(X1''), mark(X2'')))
MARK(add(0, X2)) -> A_{_ADD}(0, mark(X2))
MARK(add(len(X'), X2)) -> A_{_ADD}(a_{_len}(mark(X')), mark(X2))
MARK(add(add(X1'', X2''), X2)) -> A_{_ADD}(a_{_add}(mark(X1''), mark(X2'')), mark(X2))
MARK(add(from(X'), X2)) -> A_{_ADD}(a_{_from}(mark(X')), mark(X2))
A_{_ADD}(0, X) -> MARK(X)
MARK(add(fst(X1'', X2''), X2)) -> A_{_ADD}(a_{_fst}(mark(X1''), mark(X2'')), mark(X2))
MARK(fst(X1, len(X'))) -> A_{_FST}(mark(X1), a_{_len}(mark(X')))
MARK(fst(X1, add(X1'', X2''))) -> A_{_FST}(mark(X1), a_{_add}(mark(X1''), mark(X2'')))
MARK(fst(X1, from(X'))) -> A_{_FST}(mark(X1), a_{_from}(mark(X')))
MARK(fst(X1, fst(X1'', X2''))) -> A_{_FST}(mark(X1), a_{_fst}(mark(X1''), mark(X2'')))
MARK(fst(s(X'), X2)) -> A_{_FST}(s(X'), mark(X2))
MARK(fst(len(X'), X2)) -> A_{_FST}(a_{_len}(mark(X')), mark(X2))
MARK(fst(add(X1'', X2''), X2)) -> A_{_FST}(a_{_add}(mark(X1''), mark(X2'')), mark(X2))
MARK(fst(from(X'), X2)) -> A_{_FST}(a_{_from}(mark(X')), mark(X2))
MARK(fst(fst(X1'', X2''), X2)) -> A_{_FST}(a_{_fst}(mark(X1''), mark(X2'')), mark(X2))
MARK(cons(X1, X2)) -> MARK(X1)
MARK(len(X)) -> MARK(X)
MARK(add(X1, X2)) -> MARK(X2)
MARK(add(X1, X2)) -> MARK(X1)
MARK(from(X)) -> MARK(X)
A_{_FROM}(X) -> MARK(X)
MARK(from(X)) -> A_{_FROM}(mark(X))
MARK(fst(X1, X2)) -> MARK(X2)
MARK(fst(X1, X2)) -> MARK(X1)
A_{_FST}(s(X), cons(Y, Z)) -> MARK(Y)
MARK(fst(X1, cons(X1'', X2''))) -> A_{_FST}(mark(X1), cons(mark(X1''), X2''))

Rules:

a_{_fst}(0, Z) -> nil
a_{_fst}(s(X), cons(Y, Z)) -> cons(mark(Y), fst(X, Z))
a_{_fst}(X1, X2) -> fst(X1, X2)
a_{_from}(X) -> cons(mark(X), from(s(X)))
a_{_from}(X) -> from(X)
a_{_add}(0, X) -> mark(X)
a_{_add}(s(X), Y) -> s(add(X, Y))
a_{_add}(X1, X2) -> add(X1, X2)
a_{_len}(nil) -> 0
a_{_len}(cons(X, Z)) -> s(len(Z))
a_{_len}(X) -> len(X)
mark(fst(X1, X2)) -> a_{_fst}(mark(X1), mark(X2))
mark(from(X)) -> a_{_from}(mark(X))
mark(add(X1, X2)) -> a_{_add}(mark(X1), mark(X2))
mark(len(X)) -> a_{_len}(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)

Termination of R could not be shown.
Duration:
0:24 minutes