Term Rewriting System R:
[X, Y, Z, X1, X2]
asel(0, cons(X, Y)) -> mark(X)
asel(s(X), cons(Y, Z)) -> asel(mark(X), mark(Z))
asel(X1, X2) -> sel(X1, X2)
aindx(nil, X) -> nil
aindx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
aindx(X1, X2) -> indx(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
asel1(0, cons(X, Y)) -> mark(X)
asel1(s(X), cons(Y, Z)) -> asel1(mark(X), mark(Z))
asel1(X1, X2) -> sel1(X1, X2)
aquote(0) -> 01
aquote(s(X)) -> s1(aquote(mark(X)))
aquote(sel(X, Y)) -> asel1(mark(X), mark(Y))
aquote(X) -> quote(X)
mark(sel(X1, X2)) -> asel(mark(X1), mark(X2))
mark(indx(X1, X2)) -> aindx(mark(X1), X2)
mark(from(X)) -> afrom(X)
mark(sel1(X1, X2)) -> asel1(mark(X1), mark(X2))
mark(quote(X)) -> aquote(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)
mark(01) -> 01
mark(s1(X)) -> s1(mark(X))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

ASEL(0, cons(X, Y)) -> MARK(X)
ASEL(s(X), cons(Y, Z)) -> ASEL(mark(X), mark(Z))
ASEL(s(X), cons(Y, Z)) -> MARK(X)
ASEL(s(X), cons(Y, Z)) -> MARK(Z)
ASEL1(0, cons(X, Y)) -> MARK(X)
ASEL1(s(X), cons(Y, Z)) -> ASEL1(mark(X), mark(Z))
ASEL1(s(X), cons(Y, Z)) -> MARK(X)
ASEL1(s(X), cons(Y, Z)) -> MARK(Z)
AQUOTE(s(X)) -> AQUOTE(mark(X))
AQUOTE(s(X)) -> MARK(X)
AQUOTE(dbl(X)) -> MARK(X)
AQUOTE(sel(X, Y)) -> ASEL1(mark(X), mark(Y))
AQUOTE(sel(X, Y)) -> MARK(X)
AQUOTE(sel(X, Y)) -> MARK(Y)
MARK(dbl(X)) -> MARK(X)
MARK(dbls(X)) -> MARK(X)
MARK(sel(X1, X2)) -> ASEL(mark(X1), mark(X2))
MARK(sel(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(indx(X1, X2)) -> AINDX(mark(X1), X2)
MARK(indx(X1, X2)) -> MARK(X1)
MARK(from(X)) -> AFROM(X)
MARK(dbl1(X)) -> MARK(X)
MARK(sel1(X1, X2)) -> ASEL1(mark(X1), mark(X2))
MARK(sel1(X1, X2)) -> MARK(X1)
MARK(sel1(X1, X2)) -> MARK(X2)
MARK(quote(X)) -> AQUOTE(mark(X))
MARK(quote(X)) -> MARK(X)
MARK(s1(X)) -> MARK(X)

Furthermore, R contains one SCC.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Remaining Obligation(s)`

The following remains to be proven:
Dependency Pairs:

ASEL(s(X), cons(Y, Z)) -> MARK(Z)
AQUOTE(sel(X, Y)) -> MARK(Y)
AQUOTE(sel(X, Y)) -> MARK(X)
ASEL1(s(X), cons(Y, Z)) -> MARK(Z)
ASEL1(s(X), cons(Y, Z)) -> MARK(X)
ASEL1(s(X), cons(Y, Z)) -> ASEL1(mark(X), mark(Z))
AQUOTE(sel(X, Y)) -> ASEL1(mark(X), mark(Y))
AQUOTE(dbl(X)) -> MARK(X)
MARK(s1(X)) -> MARK(X)
MARK(quote(X)) -> MARK(X)
AQUOTE(s(X)) -> MARK(X)
AQUOTE(s(X)) -> AQUOTE(mark(X))
MARK(quote(X)) -> AQUOTE(mark(X))
MARK(sel1(X1, X2)) -> MARK(X2)
MARK(sel1(X1, X2)) -> MARK(X1)
ASEL1(0, cons(X, Y)) -> MARK(X)
MARK(sel1(X1, X2)) -> ASEL1(mark(X1), mark(X2))
MARK(dbl1(X)) -> MARK(X)
MARK(indx(X1, X2)) -> MARK(X1)
MARK(sel(X1, X2)) -> MARK(X2)
MARK(sel(X1, X2)) -> MARK(X1)
ASEL(s(X), cons(Y, Z)) -> MARK(X)
ASEL(s(X), cons(Y, Z)) -> ASEL(mark(X), mark(Z))
MARK(sel(X1, X2)) -> ASEL(mark(X1), mark(X2))
MARK(dbls(X)) -> MARK(X)
MARK(dbl(X)) -> MARK(X)
ASEL(0, cons(X, Y)) -> MARK(X)

Rules:

asel(0, cons(X, Y)) -> mark(X)
asel(s(X), cons(Y, Z)) -> asel(mark(X), mark(Z))
asel(X1, X2) -> sel(X1, X2)
aindx(nil, X) -> nil
aindx(cons(X, Y), Z) -> cons(sel(X, Z), indx(Y, Z))
aindx(X1, X2) -> indx(X1, X2)
afrom(X) -> cons(X, from(s(X)))
afrom(X) -> from(X)
asel1(0, cons(X, Y)) -> mark(X)
asel1(s(X), cons(Y, Z)) -> asel1(mark(X), mark(Z))
asel1(X1, X2) -> sel1(X1, X2)
aquote(0) -> 01
aquote(s(X)) -> s1(aquote(mark(X)))
aquote(sel(X, Y)) -> asel1(mark(X), mark(Y))
aquote(X) -> quote(X)
mark(sel(X1, X2)) -> asel(mark(X1), mark(X2))
mark(indx(X1, X2)) -> aindx(mark(X1), X2)
mark(from(X)) -> afrom(X)
mark(sel1(X1, X2)) -> asel1(mark(X1), mark(X2))
mark(quote(X)) -> aquote(mark(X))
mark(0) -> 0
mark(s(X)) -> s(X)
mark(nil) -> nil
mark(cons(X1, X2)) -> cons(X1, X2)
mark(01) -> 01
mark(s1(X)) -> s1(mark(X))

Strategy:

innermost

The Proof could not be continued due to a Timeout.
Innermost Termination of R could not be shown.
Duration:
1:01 minutes