Termination Proof using AProVE

Term Rewriting System R:
[X, L, X1, X2]
a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_INCR}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_NATS} -> A_{_ADX}(a_{_zeros})
A_{_NATS} -> A_{_ZEROS}
A_{_HEAD}(cons(X, L)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(adx(X)) -> MARK(X)
MARK(nats) -> A_{_NATS}
MARK(zeros) -> A_{_ZEROS}
MARK(head(X)) -> A_{_HEAD}(mark(X))
MARK(head(X)) -> MARK(X)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(tail(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(s(X)) -> MARK(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(head(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))
A_{_NATS} -> A_{_ADX}(a_{_zeros})
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
A_{_INCR}(cons(X, L)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

A_{_NATS} -> A_{_ADX}(a_{_zeros})

two new Dependency Pairs are created:

A_{_NATS} -> A_{_ADX}(cons(0, zeros))
A_{_NATS} -> A_{_ADX}(zeros)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(head(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))
A_{_NATS} -> A_{_ADX}(cons(0, zeros))
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(s(X)) -> MARK(X)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

MARK(incr(X)) -> A_{_INCR}(mark(X))

10 new Dependency Pairs are created:

MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(nil)) -> A_{_INCR}(nil)
MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))
MARK(incr(s(X''))) -> A_{_INCR}(s(mark(X'')))
MARK(incr(0)) -> A_{_INCR}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 3

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(s(X)) -> MARK(X)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(head(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_NATS} -> A_{_ADX}(cons(0, zeros))
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

MARK(adx(X)) -> A_{_ADX}(mark(X))

10 new Dependency Pairs are created:

MARK(adx(incr(X''))) -> A_{_ADX}(a_{_incr}(mark(X'')))
MARK(adx(adx(X''))) -> A_{_ADX}(a_{_adx}(mark(X'')))
MARK(adx(nats)) -> A_{_ADX}(a_{_nats})
MARK(adx(zeros)) -> A_{_ADX}(a_{_zeros})
MARK(adx(head(X''))) -> A_{_ADX}(a_{_head}(mark(X'')))
MARK(adx(tail(X''))) -> A_{_ADX}(a_{_tail}(mark(X'')))
MARK(adx(nil)) -> A_{_ADX}(nil)
MARK(adx(cons(X1', X2'))) -> A_{_ADX}(cons(mark(X1'), X2'))
MARK(adx(s(X''))) -> A_{_ADX}(s(mark(X'')))
MARK(adx(0)) -> A_{_ADX}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(adx(cons(X1', X2'))) -> A_{_ADX}(cons(mark(X1'), X2'))
MARK(adx(tail(X''))) -> A_{_ADX}(a_{_tail}(mark(X'')))
MARK(adx(head(X''))) -> A_{_ADX}(a_{_head}(mark(X'')))
MARK(adx(zeros)) -> A_{_ADX}(a_{_zeros})
MARK(adx(nats)) -> A_{_ADX}(a_{_nats})
MARK(adx(adx(X''))) -> A_{_ADX}(a_{_adx}(mark(X'')))
MARK(adx(incr(X''))) -> A_{_ADX}(a_{_incr}(mark(X'')))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(head(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(X)) -> A_{_HEAD}(mark(X))
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
A_{_NATS} -> A_{_ADX}(cons(0, zeros))
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

MARK(head(X)) -> A_{_HEAD}(mark(X))

10 new Dependency Pairs are created:

MARK(head(incr(X''))) -> A_{_HEAD}(a_{_incr}(mark(X'')))
MARK(head(adx(X''))) -> A_{_HEAD}(a_{_adx}(mark(X'')))
MARK(head(nats)) -> A_{_HEAD}(a_{_nats})
MARK(head(zeros)) -> A_{_HEAD}(a_{_zeros})
MARK(head(head(X''))) -> A_{_HEAD}(a_{_head}(mark(X'')))
MARK(head(tail(X''))) -> A_{_HEAD}(a_{_tail}(mark(X'')))
MARK(head(nil)) -> A_{_HEAD}(nil)
MARK(head(cons(X1', X2'))) -> A_{_HEAD}(cons(mark(X1'), X2'))
MARK(head(s(X''))) -> A_{_HEAD}(s(mark(X'')))
MARK(head(0)) -> A_{_HEAD}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(head(cons(X1', X2'))) -> A_{_HEAD}(cons(mark(X1'), X2'))
MARK(head(tail(X''))) -> A_{_HEAD}(a_{_tail}(mark(X'')))
MARK(head(head(X''))) -> A_{_HEAD}(a_{_head}(mark(X'')))
MARK(head(zeros)) -> A_{_HEAD}(a_{_zeros})
MARK(head(nats)) -> A_{_HEAD}(a_{_nats})
MARK(head(adx(X''))) -> A_{_HEAD}(a_{_adx}(mark(X'')))
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(incr(X''))) -> A_{_HEAD}(a_{_incr}(mark(X'')))
MARK(adx(tail(X''))) -> A_{_ADX}(a_{_tail}(mark(X'')))
MARK(adx(head(X''))) -> A_{_ADX}(a_{_head}(mark(X'')))
MARK(adx(zeros)) -> A_{_ADX}(a_{_zeros})
MARK(adx(nats)) -> A_{_ADX}(a_{_nats})
MARK(adx(adx(X''))) -> A_{_ADX}(a_{_adx}(mark(X'')))
MARK(adx(incr(X''))) -> A_{_ADX}(a_{_incr}(mark(X'')))
MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(tail(X)) -> A_{_TAIL}(mark(X))
MARK(head(X)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> MARK(X)
A_{_NATS} -> A_{_ADX}(cons(0, zeros))
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
MARK(adx(cons(X1', X2'))) -> A_{_ADX}(cons(mark(X1'), X2'))

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

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

MARK(tail(X)) -> A_{_TAIL}(mark(X))

10 new Dependency Pairs are created:

MARK(tail(incr(X''))) -> A_{_TAIL}(a_{_incr}(mark(X'')))
MARK(tail(adx(X''))) -> A_{_TAIL}(a_{_adx}(mark(X'')))
MARK(tail(nats)) -> A_{_TAIL}(a_{_nats})
MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})
MARK(tail(head(X''))) -> A_{_TAIL}(a_{_head}(mark(X'')))
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
MARK(tail(nil)) -> A_{_TAIL}(nil)
MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(tail(s(X''))) -> A_{_TAIL}(s(mark(X'')))
MARK(tail(0)) -> A_{_TAIL}(0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(tail(cons(X1', X2'))) -> A_{_TAIL}(cons(mark(X1'), X2'))
MARK(tail(tail(X''))) -> A_{_TAIL}(a_{_tail}(mark(X'')))
MARK(tail(head(X''))) -> A_{_TAIL}(a_{_head}(mark(X'')))
MARK(tail(zeros)) -> A_{_TAIL}(a_{_zeros})
MARK(tail(nats)) -> A_{_TAIL}(a_{_nats})
MARK(tail(adx(X''))) -> A_{_TAIL}(a_{_adx}(mark(X'')))
A_{_TAIL}(cons(X, L)) -> MARK(L)
MARK(tail(incr(X''))) -> A_{_TAIL}(a_{_incr}(mark(X'')))
MARK(head(tail(X''))) -> A_{_HEAD}(a_{_tail}(mark(X'')))
MARK(head(head(X''))) -> A_{_HEAD}(a_{_head}(mark(X'')))
MARK(head(zeros)) -> A_{_HEAD}(a_{_zeros})
MARK(head(nats)) -> A_{_HEAD}(a_{_nats})
MARK(head(adx(X''))) -> A_{_HEAD}(a_{_adx}(mark(X'')))
MARK(head(incr(X''))) -> A_{_HEAD}(a_{_incr}(mark(X'')))
MARK(adx(cons(X1', X2'))) -> A_{_ADX}(cons(mark(X1'), X2'))
MARK(adx(tail(X''))) -> A_{_ADX}(a_{_tail}(mark(X'')))
MARK(adx(head(X''))) -> A_{_ADX}(a_{_head}(mark(X'')))
MARK(adx(zeros)) -> A_{_ADX}(a_{_zeros})
MARK(adx(nats)) -> A_{_ADX}(a_{_nats})
MARK(adx(adx(X''))) -> A_{_ADX}(a_{_adx}(mark(X'')))
A_{_ADX}(cons(X, L)) -> MARK(X)
MARK(adx(incr(X''))) -> A_{_ADX}(a_{_incr}(mark(X'')))
MARK(incr(cons(X1', X2'))) -> A_{_INCR}(cons(mark(X1'), X2'))
MARK(incr(tail(X''))) -> A_{_INCR}(a_{_tail}(mark(X'')))
MARK(incr(head(X''))) -> A_{_INCR}(a_{_head}(mark(X'')))
MARK(incr(zeros)) -> A_{_INCR}(a_{_zeros})
MARK(incr(nats)) -> A_{_INCR}(a_{_nats})
MARK(incr(adx(X''))) -> A_{_INCR}(a_{_adx}(mark(X'')))
MARK(incr(incr(X''))) -> A_{_INCR}(a_{_incr}(mark(X'')))
MARK(s(X)) -> MARK(X)
MARK(cons(X1, X2)) -> MARK(X1)
MARK(tail(X)) -> MARK(X)
MARK(head(X)) -> MARK(X)
A_{_INCR}(cons(X, L)) -> MARK(X)
A_{_ADX}(cons(X, L)) -> A_{_INCR}(cons(mark(X), adx(L)))
A_{_NATS} -> A_{_ADX}(cons(0, zeros))
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
A_{_HEAD}(cons(X, L)) -> MARK(X)
MARK(head(cons(X1', X2'))) -> A_{_HEAD}(cons(mark(X1'), X2'))

Rules:

a_{_incr}(nil) -> nil
a_{_incr}(cons(X, L)) -> cons(s(mark(X)), incr(L))
a_{_incr}(X) -> incr(X)
a_{_adx}(nil) -> nil
a_{_adx}(cons(X, L)) -> a_{_incr}(cons(mark(X), adx(L)))
a_{_adx}(X) -> adx(X)
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_head}(cons(X, L)) -> mark(X)
a_{_head}(X) -> head(X)
a_{_tail}(cons(X, L)) -> mark(L)
a_{_tail}(X) -> tail(X)
mark(incr(X)) -> a_{_incr}(mark(X))
mark(adx(X)) -> a_{_adx}(mark(X))
mark(nats) -> a_{_nats}
mark(zeros) -> a_{_zeros}
mark(head(X)) -> a_{_head}(mark(X))
mark(tail(X)) -> a_{_tail}(mark(X))
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(mark(X1), X2)
mark(s(X)) -> s(mark(X))
mark(0) -> 0

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:27 minutes