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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP

(0) Obligation:

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

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → 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:

INCR(cons(X, L)) → ACTIVATE(L)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)
NATS → ADX(zeros)
NATS → ZEROS
TAIL(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__adx(X)) → ADX(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → 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 4 less nodes.

(4) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ADX(cons(X, L)) → ACTIVATE(L)

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → 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.


ADX(cons(X, L)) → ACTIVATE(L)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__incr(x1)  =  x1
INCR(x1)  =  x1
cons(x1, x2)  =  x2
n__adx(x1)  =  n__adx(x1)
ADX(x1)  =  ADX(x1)
activate(x1)  =  x1
incr(x1)  =  x1
adx(x1)  =  adx(x1)
n__zeros  =  n__zeros
zeros  =  zeros
s(x1)  =  s(x1)
nil  =  nil
0  =  0

Lexicographic Path Order [LPO].
Precedence:
[nzeros, zeros] > 0 > [nadx1, ADX1, adx1, s1, nil]


The following usable rules [FROCOS05] were oriented:

activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
incr(nil) → nil
incr(X) → n__incr(X)
adx(nil) → nil
adx(X) → n__adx(X)
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeros → cons(0, n__zeros)
zeros → n__zeros

(6) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(X)
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))

The TRS R consists of the following rules:

incr(nil) → nil
incr(cons(X, L)) → cons(s(X), n__incr(activate(L)))
adx(nil) → nil
adx(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
nats → adx(zeros)
zeros → cons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → 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.


ACTIVATE(n__incr(X)) → INCR(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  x1
n__incr(x1)  =  n__incr(x1)
INCR(x1)  =  x1
cons(x1, x2)  =  x2
n__adx(x1)  =  n__adx
ADX(x1)  =  ADX
activate(x1)  =  activate(x1)
incr(x1)  =  incr(x1)
adx(x1)  =  adx
n__zeros  =  n__zeros
zeros  =  zeros
s(x1)  =  s
nil  =  nil
0  =  0

Lexicographic Path Order [LPO].
Precedence:
activate1 > [nincr1, nadx, ADX, incr1, adx, s, nil]
zeros > nzeros > [nincr1, nadx, ADX, incr1, adx, s, nil]
zeros > 0 > [nincr1, nadx, ADX, incr1, adx, s, nil]


The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))

The TRS R consists of the following rules:

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

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