Termination Proof using AProVE

Term Rewriting System R:
[X, Y, X1, X2]
a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_incr}(cons(X, Y)) -> cons(s(X), incr(Y))
a_{_incr}(X) -> incr(X)
a_{_adx}(cons(X, Y)) -> a_{_incr}(cons(X, adx(Y)))
a_{_adx}(X) -> adx(X)
a_{_hd}(cons(X, Y)) -> mark(X)
a_{_hd}(X) -> hd(X)
a_{_tl}(cons(X, Y)) -> mark(Y)
a_{_tl}(X) -> tl(X)
mark(nats) -> a_{_nats}
mark(adx(X)) -> a_{_adx}(mark(X))
mark(zeros) -> a_{_zeros}
mark(incr(X)) -> a_{_incr}(mark(X))
mark(hd(X)) -> a_{_hd}(mark(X))
mark(tl(X)) -> a_{_tl}(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

A_{_NATS} -> A_{_ADX}(a_{_zeros})
A_{_NATS} -> A_{_ZEROS}
A_{_ADX}(cons(X, Y)) -> A_{_INCR}(cons(X, adx(Y)))
A_{_HD}(cons(X, Y)) -> MARK(X)
A_{_TL}(cons(X, Y)) -> MARK(Y)
MARK(nats) -> A_{_NATS}
MARK(adx(X)) -> A_{_ADX}(mark(X))
MARK(adx(X)) -> MARK(X)
MARK(zeros) -> A_{_ZEROS}
MARK(incr(X)) -> A_{_INCR}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(hd(X)) -> A_{_HD}(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(tl(X)) -> A_{_TL}(mark(X))
MARK(tl(X)) -> MARK(X)

Furthermore, R contains one SCC.

     ↳DPs

       →DP Problem 1

         ↳Narrowing Transformation

Dependency Pairs:

MARK(tl(X)) -> MARK(X)
A_{_TL}(cons(X, Y)) -> MARK(Y)
MARK(tl(X)) -> A_{_TL}(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(hd(X)) -> A_{_HD}(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
A_{_HD}(cons(X, Y)) -> MARK(X)

Rules:

a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_incr}(cons(X, Y)) -> cons(s(X), incr(Y))
a_{_incr}(X) -> incr(X)
a_{_adx}(cons(X, Y)) -> a_{_incr}(cons(X, adx(Y)))
a_{_adx}(X) -> adx(X)
a_{_hd}(cons(X, Y)) -> mark(X)
a_{_hd}(X) -> hd(X)
a_{_tl}(cons(X, Y)) -> mark(Y)
a_{_tl}(X) -> tl(X)
mark(nats) -> a_{_nats}
mark(adx(X)) -> a_{_adx}(mark(X))
mark(zeros) -> a_{_zeros}
mark(incr(X)) -> a_{_incr}(mark(X))
mark(hd(X)) -> a_{_hd}(mark(X))
mark(tl(X)) -> a_{_tl}(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)

Strategy:

innermost

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

MARK(hd(X)) -> A_{_HD}(mark(X))

nine new Dependency Pairs are created:

MARK(hd(nats)) -> A_{_HD}(a_{_nats})
MARK(hd(adx(X''))) -> A_{_HD}(a_{_adx}(mark(X'')))
MARK(hd(zeros)) -> A_{_HD}(a_{_zeros})
MARK(hd(incr(X''))) -> A_{_HD}(a_{_incr}(mark(X'')))
MARK(hd(hd(X''))) -> A_{_HD}(a_{_hd}(mark(X'')))
MARK(hd(tl(X''))) -> A_{_HD}(a_{_tl}(mark(X'')))
MARK(hd(cons(X1', X2'))) -> A_{_HD}(cons(X1', X2'))
MARK(hd(0)) -> A_{_HD}(0)
MARK(hd(s(X''))) -> A_{_HD}(s(X''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Nar

           →DP Problem 2

             ↳Narrowing Transformation

Dependency Pairs:

MARK(hd(cons(X1', X2'))) -> A_{_HD}(cons(X1', X2'))
MARK(hd(tl(X''))) -> A_{_HD}(a_{_tl}(mark(X'')))
MARK(hd(hd(X''))) -> A_{_HD}(a_{_hd}(mark(X'')))
MARK(hd(incr(X''))) -> A_{_HD}(a_{_incr}(mark(X'')))
MARK(hd(zeros)) -> A_{_HD}(a_{_zeros})
MARK(hd(adx(X''))) -> A_{_HD}(a_{_adx}(mark(X'')))
A_{_HD}(cons(X, Y)) -> MARK(X)
MARK(hd(nats)) -> A_{_HD}(a_{_nats})
A_{_TL}(cons(X, Y)) -> MARK(Y)
MARK(tl(X)) -> A_{_TL}(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
MARK(tl(X)) -> MARK(X)

Rules:

a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_incr}(cons(X, Y)) -> cons(s(X), incr(Y))
a_{_incr}(X) -> incr(X)
a_{_adx}(cons(X, Y)) -> a_{_incr}(cons(X, adx(Y)))
a_{_adx}(X) -> adx(X)
a_{_hd}(cons(X, Y)) -> mark(X)
a_{_hd}(X) -> hd(X)
a_{_tl}(cons(X, Y)) -> mark(Y)
a_{_tl}(X) -> tl(X)
mark(nats) -> a_{_nats}
mark(adx(X)) -> a_{_adx}(mark(X))
mark(zeros) -> a_{_zeros}
mark(incr(X)) -> a_{_incr}(mark(X))
mark(hd(X)) -> a_{_hd}(mark(X))
mark(tl(X)) -> a_{_tl}(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)

Strategy:

innermost

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

MARK(tl(X)) -> A_{_TL}(mark(X))

nine new Dependency Pairs are created:

MARK(tl(nats)) -> A_{_TL}(a_{_nats})
MARK(tl(adx(X''))) -> A_{_TL}(a_{_adx}(mark(X'')))
MARK(tl(zeros)) -> A_{_TL}(a_{_zeros})
MARK(tl(incr(X''))) -> A_{_TL}(a_{_incr}(mark(X'')))
MARK(tl(hd(X''))) -> A_{_TL}(a_{_hd}(mark(X'')))
MARK(tl(tl(X''))) -> A_{_TL}(a_{_tl}(mark(X'')))
MARK(tl(cons(X1', X2'))) -> A_{_TL}(cons(X1', X2'))
MARK(tl(0)) -> A_{_TL}(0)
MARK(tl(s(X''))) -> A_{_TL}(s(X''))

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(tl(cons(X1', X2'))) -> A_{_TL}(cons(X1', X2'))
MARK(tl(tl(X''))) -> A_{_TL}(a_{_tl}(mark(X'')))
MARK(tl(hd(X''))) -> A_{_TL}(a_{_hd}(mark(X'')))
MARK(tl(incr(X''))) -> A_{_TL}(a_{_incr}(mark(X'')))
MARK(tl(zeros)) -> A_{_TL}(a_{_zeros})
MARK(tl(adx(X''))) -> A_{_TL}(a_{_adx}(mark(X'')))
A_{_TL}(cons(X, Y)) -> MARK(Y)
MARK(tl(nats)) -> A_{_TL}(a_{_nats})
MARK(hd(tl(X''))) -> A_{_HD}(a_{_tl}(mark(X'')))
MARK(hd(hd(X''))) -> A_{_HD}(a_{_hd}(mark(X'')))
MARK(hd(incr(X''))) -> A_{_HD}(a_{_incr}(mark(X'')))
MARK(hd(zeros)) -> A_{_HD}(a_{_zeros})
MARK(hd(adx(X''))) -> A_{_HD}(a_{_adx}(mark(X'')))
MARK(hd(nats)) -> A_{_HD}(a_{_nats})
MARK(tl(X)) -> MARK(X)
MARK(hd(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
A_{_HD}(cons(X, Y)) -> MARK(X)
MARK(hd(cons(X1', X2'))) -> A_{_HD}(cons(X1', X2'))

Rules:

a_{_nats} -> a_{_adx}(a_{_zeros})
a_{_nats} -> nats
a_{_zeros} -> cons(0, zeros)
a_{_zeros} -> zeros
a_{_incr}(cons(X, Y)) -> cons(s(X), incr(Y))
a_{_incr}(X) -> incr(X)
a_{_adx}(cons(X, Y)) -> a_{_incr}(cons(X, adx(Y)))
a_{_adx}(X) -> adx(X)
a_{_hd}(cons(X, Y)) -> mark(X)
a_{_hd}(X) -> hd(X)
a_{_tl}(cons(X, Y)) -> mark(Y)
a_{_tl}(X) -> tl(X)
mark(nats) -> a_{_nats}
mark(adx(X)) -> a_{_adx}(mark(X))
mark(zeros) -> a_{_zeros}
mark(incr(X)) -> a_{_incr}(mark(X))
mark(hd(X)) -> a_{_hd}(mark(X))
mark(tl(X)) -> a_{_tl}(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:06 minutes