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(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 remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x0, x1)
A__HD(x0, x1)  =  A__HD(x0)
A__TL(x0, x1)  =  A__TL(x1)

Tags:
MARK has argument tags [1,0] and root tag 1
A__HD has argument tags [1,6] and root tag 0
A__TL has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MARK(x1)  =  x1
adx(x1)  =  x1
incr(x1)  =  x1
hd(x1)  =  hd(x1)
A__HD(x1)  =  x1
mark(x1)  =  mark(x1)
cons(x1, x2)  =  cons(x1, x2)
tl(x1)  =  tl(x1)
A__TL(x1)  =  A__TL
nats  =  nats
a__nats  =  a__nats
a__adx(x1)  =  x1
zeros  =  zeros
a__zeros  =  a__zeros
a__incr(x1)  =  x1
a__hd(x1)  =  a__hd(x1)
a__tl(x1)  =  a__tl(x1)
0  =  0
s(x1)  =  x1

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

[hd₁, mark₁, tl₁, a_hd1, a_tl1] > a_nats > nats
[hd₁, mark₁, tl₁, a_hd1, a_tl1] > a_nats > a_zeros > cons₂
[hd₁, mark₁, tl₁, a_hd1, a_tl1] > a_nats > a_zeros > zeros
[hd₁, mark₁, tl₁, a_hd1, a_tl1] > a_nats > a_zeros > 0

Status:

hd₁: [1]
mark₁: [1]
cons₂: [2,1]
tl₁: [1]
A_TL: []
nats: []
a_nats: []
zeros: []
a_zeros: []
a_hd1: [1]
a_tl1: [1]
0: []

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

MARK(adx(X)) → MARK(X)
MARK(incr(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.

(7) 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)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)

Tags:
MARK has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MARK(x1)  =  MARK
adx(x1)  =  adx(x1)
incr(x1)  =  x1

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

trivial

Status:

MARK: []
adx₁: [1]

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

MARK(incr(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.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(incr(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MARK(x0, x1)  =  MARK(x1)

Tags:
MARK has argument tags [1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
MARK(x1)  =  MARK
incr(x1)  =  incr(x1)

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

trivial

Status:

MARK: []
incr₁: [1]

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

Q DP problem:
P is empty.
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.

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE