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(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(x1)
A__TL(x0, x1)  =  A__TL(x0, x1)

Tags:
MARK has argument tags [2,2] and root tag 0
A__HD has argument tags [5,2] and root tag 0
A__TL has argument tags [1,5] and root tag 3

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__HD(x₁)) = 1   
POL(A__TL(x₁)) = x₁   
POL(MARK(x₁)) = 0   
POL(a__adx(x₁)) = x₁   
POL(a__hd(x₁)) = x₁   
POL(a__incr(x₁)) = x₁   
POL(a__nats) = 1   
POL(a__tl(x₁)) = 1 + x₁   
POL(a__zeros) = 1   
POL(adx(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(hd(x₁)) = x₁   
POL(incr(x₁)) = x₁   
POL(mark(x₁)) = x₁   
POL(nats) = 1   
POL(s(x₁)) = x₁   
POL(tl(x₁)) = 1 + x₁   
POL(zeros) = 1   

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)
MARK(hd(X)) → A__HD(mark(X))
A__HD(cons(X, Y)) → MARK(X)
MARK(hd(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(hd(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)
A__HD(x0, x1)  =  A__HD(x0)

Tags:
MARK has argument tags [2,1] and root tag 0
A__HD has argument tags [1,2] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__HD(x₁)) = x₁   
POL(MARK(x₁)) = 1   
POL(a__adx(x₁)) = x₁   
POL(a__hd(x₁)) = 1 + x₁   
POL(a__incr(x₁)) = x₁   
POL(a__nats) = 1   
POL(a__tl(x₁)) = 1 + x₁   
POL(a__zeros) = 1   
POL(adx(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(hd(x₁)) = 1 + x₁   
POL(incr(x₁)) = x₁   
POL(mark(x₁)) = 1 + x₁   
POL(nats) = 0   
POL(s(x₁)) = 0   
POL(tl(x₁)) = 1 + x₁   
POL(zeros) = 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

(8) 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)

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(hd(X)) → A__HD(mark(X))
A__HD(cons(X, Y)) → 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)
A__HD(x0, x1)  =  A__HD(x0, x1)

Tags:
MARK has argument tags [2,1] and root tag 1
A__HD has argument tags [0,1] and root tag 0

Comparison: MS
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(A__HD(x₁)) = 1   
POL(MARK(x₁)) = 1   
POL(a__adx(x₁)) = x₁   
POL(a__hd(x₁)) = 1 + x₁   
POL(a__incr(x₁)) = x₁   
POL(a__nats) = 0   
POL(a__tl(x₁)) = 1 + x₁   
POL(a__zeros) = 0   
POL(adx(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(hd(x₁)) = 1 + x₁   
POL(incr(x₁)) = x₁   
POL(mark(x₁)) = x₁   
POL(nats) = 0   
POL(s(x₁)) = x₁   
POL(tl(x₁)) = 1 + x₁   
POL(zeros) = 0   

The following usable rules [FROCOS05] were oriented: none

(10) 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.

(11) 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 [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(MARK(x₁)) = 1   
POL(adx(x₁)) = x₁   
POL(incr(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

MARK(adx(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.

(13) 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:
Polynomial interpretation [POLO]:

POL(MARK(x₁)) = 1   
POL(adx(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(14) 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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE