Term Rewriting System R:
[X, Y, X1, X2]
anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(mark(X))
mark(cons(X1, X2)) -> cons(X1, X2)
mark(0) -> 0
mark(s(X)) -> s(X)

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

ANATS -> AADX(azeros)
ANATS -> AZEROS
AADX(cons(X, Y)) -> AINCR(cons(X, adx(Y)))
AHD(cons(X, Y)) -> MARK(X)
ATL(cons(X, Y)) -> MARK(Y)
MARK(nats) -> ANATS
MARK(adx(X)) -> AADX(mark(X))
MARK(adx(X)) -> MARK(X)
MARK(zeros) -> AZEROS
MARK(incr(X)) -> AINCR(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(hd(X)) -> AHD(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(tl(X)) -> ATL(mark(X))
MARK(tl(X)) -> MARK(X)

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Narrowing Transformation


Dependency Pairs:

MARK(tl(X)) -> MARK(X)
ATL(cons(X, Y)) -> MARK(Y)
MARK(tl(X)) -> ATL(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(hd(X)) -> AHD(mark(X))
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
AHD(cons(X, Y)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(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)) -> AHD(mark(X))
nine new Dependency Pairs are created:

MARK(hd(nats)) -> AHD(anats)
MARK(hd(adx(X''))) -> AHD(aadx(mark(X'')))
MARK(hd(zeros)) -> AHD(azeros)
MARK(hd(incr(X''))) -> AHD(aincr(mark(X'')))
MARK(hd(hd(X''))) -> AHD(ahd(mark(X'')))
MARK(hd(tl(X''))) -> AHD(atl(mark(X'')))
MARK(hd(cons(X1', X2'))) -> AHD(cons(X1', X2'))
MARK(hd(0)) -> AHD(0)
MARK(hd(s(X''))) -> AHD(s(X''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
           →DP Problem 2
Narrowing Transformation


Dependency Pairs:

MARK(hd(cons(X1', X2'))) -> AHD(cons(X1', X2'))
MARK(hd(tl(X''))) -> AHD(atl(mark(X'')))
MARK(hd(hd(X''))) -> AHD(ahd(mark(X'')))
MARK(hd(incr(X''))) -> AHD(aincr(mark(X'')))
MARK(hd(zeros)) -> AHD(azeros)
MARK(hd(adx(X''))) -> AHD(aadx(mark(X'')))
AHD(cons(X, Y)) -> MARK(X)
MARK(hd(nats)) -> AHD(anats)
ATL(cons(X, Y)) -> MARK(Y)
MARK(tl(X)) -> ATL(mark(X))
MARK(hd(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
MARK(tl(X)) -> MARK(X)


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(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)) -> ATL(mark(X))
nine new Dependency Pairs are created:

MARK(tl(nats)) -> ATL(anats)
MARK(tl(adx(X''))) -> ATL(aadx(mark(X'')))
MARK(tl(zeros)) -> ATL(azeros)
MARK(tl(incr(X''))) -> ATL(aincr(mark(X'')))
MARK(tl(hd(X''))) -> ATL(ahd(mark(X'')))
MARK(tl(tl(X''))) -> ATL(atl(mark(X'')))
MARK(tl(cons(X1', X2'))) -> ATL(cons(X1', X2'))
MARK(tl(0)) -> ATL(0)
MARK(tl(s(X''))) -> ATL(s(X''))

The transformation is resulting in one new DP problem:



   R
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'))) -> ATL(cons(X1', X2'))
MARK(tl(tl(X''))) -> ATL(atl(mark(X'')))
MARK(tl(hd(X''))) -> ATL(ahd(mark(X'')))
MARK(tl(incr(X''))) -> ATL(aincr(mark(X'')))
MARK(tl(zeros)) -> ATL(azeros)
MARK(tl(adx(X''))) -> ATL(aadx(mark(X'')))
ATL(cons(X, Y)) -> MARK(Y)
MARK(tl(nats)) -> ATL(anats)
MARK(hd(tl(X''))) -> AHD(atl(mark(X'')))
MARK(hd(hd(X''))) -> AHD(ahd(mark(X'')))
MARK(hd(incr(X''))) -> AHD(aincr(mark(X'')))
MARK(hd(zeros)) -> AHD(azeros)
MARK(hd(adx(X''))) -> AHD(aadx(mark(X'')))
MARK(hd(nats)) -> AHD(anats)
MARK(tl(X)) -> MARK(X)
MARK(hd(X)) -> MARK(X)
MARK(incr(X)) -> MARK(X)
MARK(adx(X)) -> MARK(X)
AHD(cons(X, Y)) -> MARK(X)
MARK(hd(cons(X1', X2'))) -> AHD(cons(X1', X2'))


Rules:


anats -> aadx(azeros)
anats -> nats
azeros -> cons(0, zeros)
azeros -> zeros
aincr(cons(X, Y)) -> cons(s(X), incr(Y))
aincr(X) -> incr(X)
aadx(cons(X, Y)) -> aincr(cons(X, adx(Y)))
aadx(X) -> adx(X)
ahd(cons(X, Y)) -> mark(X)
ahd(X) -> hd(X)
atl(cons(X, Y)) -> mark(Y)
atl(X) -> tl(X)
mark(nats) -> anats
mark(adx(X)) -> aadx(mark(X))
mark(zeros) -> azeros
mark(incr(X)) -> aincr(mark(X))
mark(hd(X)) -> ahd(mark(X))
mark(tl(X)) -> atl(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:
0:07 minutes