Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
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(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)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
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(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)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 7 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(adx(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(hd(X)) → A__HD(mark(X))
A__HD(cons(X, Y)) → MARK(X)
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))
A__TL(cons(X, Y)) → MARK(Y)
MARK(tl(X)) → MARK(X)

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
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(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)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(adx(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
A__HD(cons(X, Y)) → MARK(X)
MARK(hd(X)) → MARK(X)
A__TL(cons(X, Y)) → MARK(Y)
MARK(tl(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1)  =  MARK(x1)
adx(x1)  =  adx(x1)
incr(x1)  =  incr(x1)
hd(x1)  =  hd(x1)
A__HD(x1)  =  A__HD(x1)
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons(x1, x2)
tl(x1)  =  tl(x1)
A__TL(x1)  =  A__TL(x1)
a__incr(x1)  =  a__incr(x1)
a__hd(x1)  =  a__hd(x1)
a__tl(x1)  =  a__tl(x1)
a__adx(x1)  =  a__adx(x1)
zeros  =  zeros
a__zeros  =  a__zeros
nats  =  nats
a__nats  =  a__nats
0  =  0
s(x1)  =  s

Lexicographic path order with status [LPO].
Quasi-Precedence:

[MARK₁, hd₁, A_HD1, mark₁, tl₁, A_TL1, a_hd1, a_tl1, nats, a_nats] > a_adx1 > adx₁ > cons₂
[MARK₁, hd₁, A_HD1, mark₁, tl₁, A_TL1, a_hd1, a_tl1, nats, a_nats] > a_adx1 > [incr₁, a_incr1] > s > cons₂
[MARK₁, hd₁, A_HD1, mark₁, tl₁, A_TL1, a_hd1, a_tl1, nats, a_nats] > a_zeros > zeros > cons₂
[MARK₁, hd₁, A_HD1, mark₁, tl₁, A_TL1, a_hd1, a_tl1, nats, a_nats] > a_zeros > 0 > cons₂

Status:

a_hd1: [1]
a_adx1: [1]
a_tl1: [1]
a_zeros: []
A_TL1: [1]
mark₁: [1]
s: []
0: []
hd₁: [1]
cons₂: [2,1]
MARK₁: [1]
a_nats: []
adx₁: [1]
incr₁: [1]
a_incr1: [1]
tl₁: [1]
zeros: []
A_HD1: [1]
nats: []

The following usable rules [FROCOS05] were oriented:

mark(cons(X1, X2)) → cons(X1, X2)
mark(incr(X)) → a__incr(mark(X))
a__hd(cons(X, Y)) → mark(X)
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
a__tl(cons(X, Y)) → mark(Y)
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(nats) → a__nats
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__nats → a__adx(a__zeros)
a__incr(X) → incr(X)
a__zeros → zeros
a__tl(X) → tl(X)
a__hd(X) → hd(X)
mark(s(X)) → s(X)
mark(0) → 0
a__adx(X) → adx(X)
a__nats → nats

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MARK(hd(X)) → A__HD(mark(X))
MARK(tl(X)) → A__TL(mark(X))

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
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(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)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(8) TRUE