Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

INCR(cons(X, L)) → INCR(L)
ADX(cons(X, L)) → INCR(cons(X, adx(L)))
ADX(cons(X, L)) → ADX(L)
NATS → ADX(zeros)
NATS → ZEROS
ZEROS → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

ZEROS → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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

(8) Obligation:

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

INCR(cons(X, L)) → INCR(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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.

INCR(cons(X, L)) → INCR(L)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
INCR(x1)  =  x1
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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

(13) Obligation:

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

ADX(cons(X, L)) → ADX(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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

(14) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADX(cons(X, L)) → ADX(L)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADX(x1)  =  x1
cons(x1, x2)  =  cons(x2)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(15) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), incr(L))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, adx(L)))
nats → adx(zeros)
zeros → cons(0, zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → L

The set Q consists of the following terms:

incr(nil)
incr(cons(x0, x1))
adx(nil)
adx(cons(x0, x1))
nats
zeros
head(cons(x0, x1))
tail(cons(x0, x1))

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE