Termination Proof using AProVE

Term Rewriting System R:
[X, Y]
nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

NATS -> ADX(zeros)
NATS -> ZEROS
ZEROS -> ZEROS
INCR(cons(X, Y)) -> INCR(Y)
ADX(cons(X, Y)) -> INCR(cons(X, adx(Y)))
ADX(cons(X, Y)) -> ADX(Y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
INCR(cons(X, Y)) -> INCR(Y)

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y
Dependency Pair:
ZEROS -> ZEROS

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y
Dependency Pair:
ADX(cons(X, Y)) -> ADX(Y)

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
INCR(cons(X, Y)) -> INCR(Y)

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y
Dependency Pair:
ZEROS -> ZEROS

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y
Dependency Pair:
ADX(cons(X, Y)) -> ADX(Y)

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pair:
INCR(cons(X, Y)) -> INCR(Y)

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y
Dependency Pair:
ZEROS -> ZEROS

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y
Dependency Pair:
ADX(cons(X, Y)) -> ADX(Y)

Rules:

nats -> adx(zeros)
zeros -> cons(0, zeros)
incr(cons(X, Y)) -> cons(s(X), incr(Y))
adx(cons(X, Y)) -> incr(cons(X, adx(Y)))
hd(cons(X, Y)) -> X
tl(cons(X, Y)) -> Y

Termination of R could not be shown.
Duration:
0:00 minutes