(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))))
natsadx(zeros)
zeroscons(0, n__zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zerosn__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(head(x1)) = 1 + 2·x1   
POL(incr(x1)) = x1   
POL(n__adx(x1)) = 2·x1   
POL(n__incr(x1)) = x1   
POL(n__zeros) = 0   
POL(nats) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 2·x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

head(cons(X, L)) → X


(2) 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))))
natsadx(zeros)
zeroscons(0, n__zeros)
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zerosn__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = 2·x1   
POL(cons(x1, x2)) = x1 + 2·x2   
POL(incr(x1)) = x1   
POL(n__adx(x1)) = 2·x1   
POL(n__incr(x1)) = x1   
POL(n__zeros) = 0   
POL(nats) = 2   
POL(nil) = 1   
POL(s(x1)) = x1   
POL(tail(x1)) = 2·x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

adx(nil) → nil
natsadx(zeros)


(4) 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(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeroscons(0, n__zeros)
tail(cons(X, L)) → activate(L)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zerosn__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(5) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(activate(x1)) = x1   
POL(adx(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(n__adx(x1)) = x1   
POL(n__incr(x1)) = x1   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(tail(x1)) = 1 + 2·x1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

tail(cons(X, L)) → activate(L)


(6) 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(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeroscons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zerosn__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(7) DependencyPairsProof (EQUIVALENT transformation)

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

(8) 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)
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ACTIVATE(n__adx(X)) → ACTIVATE(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(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeroscons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zerosn__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

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

(9) DependencyGraphProof (EQUIVALENT transformation)

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

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

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


Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(ADX(x1)) = 1 + 2·x1   
POL(INCR(x1)) = x1   
POL(activate(x1)) = x1   
POL(adx(x1)) = 1 + 2·x1   
POL(cons(x1, x2)) = x1 + x2   
POL(incr(x1)) = x1   
POL(n__adx(x1)) = 1 + 2·x1   
POL(n__incr(x1)) = x1   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(zeros) = 0   

(12) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(activate(X))
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ADX(activate(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(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeroscons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zerosn__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → 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.


ACTIVATE(n__incr(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ADX(x1) ) = 1


POL( INCR(x1) ) = x1 + 1


POL( activate(x1) ) = x1


POL( n__incr(x1) ) = x1 + 1


POL( incr(x1) ) = x1 + 1


POL( n__adx(x1) ) = 1


POL( adx(x1) ) = 1


POL( n__zeros ) = 0


POL( zeros ) = 0


POL( nil ) = 2


POL( cons(x1, x2) ) = max{0, x2 - 1}


POL( s(x1) ) = 2x1 + 2


POL( 0 ) = 0


POL( ACTIVATE(x1) ) = x1



The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(activate(X))
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(activate(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(cons(X, L)) → incr(cons(X, n__adx(activate(L))))
zeroscons(0, n__zeros)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zerosn__zeros
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(n__zeros) → zeros
activate(X) → X

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

(15) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__incr(X)) → INCR(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(x0)) → INCR(x0)

(16) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(x0)) → INCR(x0)

The TRS R consists of the following rules:

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

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

(17) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__adx(X)) → ADX(activate(X)) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(x0)) → ADX(x0)

(18) 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))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(x0)) → ADX(x0)

The TRS R consists of the following rules:

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

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

(19) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__incr(n__zeros)) → INCR(zeros) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))
ACTIVATE(n__incr(n__zeros)) → INCR(n__zeros)

(20) 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))))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))
ACTIVATE(n__incr(n__zeros)) → INCR(n__zeros)

The TRS R consists of the following rules:

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

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))

The TRS R consists of the following rules:

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

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

(23) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__adx(n__zeros)) → ADX(zeros) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(n__zeros)

(24) Obligation:

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

ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(n__zeros)

The TRS R consists of the following rules:

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

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

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))

The TRS R consists of the following rules:

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

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__incr(n__zeros)) → INCR(cons(0, n__zeros))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( ADX(x1) ) = 2


POL( INCR(x1) ) = x1 + 2


POL( activate(x1) ) = x1


POL( n__incr(x1) ) = x1 + 2


POL( incr(x1) ) = x1 + 2


POL( n__adx(x1) ) = 2


POL( adx(x1) ) = 2


POL( cons(x1, x2) ) = max{0, x2 - 2}


POL( n__zeros ) = 1


POL( zeros ) = 1


POL( nil ) = 1


POL( s(x1) ) = max{0, x1 - 1}


POL( 0 ) = 2


POL( ACTIVATE(x1) ) = x1



The following usable rules [FROCOS05] were oriented:

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

(28) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))

The TRS R consists of the following rules:

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

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = [1/2] + [1/2]x1   
POL(ADX(x1)) = [1] + [1/4]x1   
POL(INCR(x1)) = [1/2] + [1/4]x1   
POL(activate(x1)) = x1   
POL(adx(x1)) = [1] + x1   
POL(cons(x1, x2)) = [2]x2   
POL(incr(x1)) = [1/2]x1   
POL(n__adx(x1)) = [1] + x1   
POL(n__incr(x1)) = [1/2]x1   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = [1] + [1/4]x1   
POL(zeros) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [FROCOS05] were oriented:

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

(30) Obligation:

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

INCR(cons(X, L)) → ACTIVATE(L)
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros))

The TRS R consists of the following rules:

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

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

(31) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = ACTIVATE(n__adx(activate(n__zeros))) evaluates to t =ACTIVATE(n__adx(activate(n__zeros)))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]




Rewriting sequence

ACTIVATE(n__adx(activate(n__zeros)))ACTIVATE(n__adx(n__zeros))
with rule activate(X) → X at position [0,0] and matcher [X / n__zeros]

ACTIVATE(n__adx(n__zeros))ADX(cons(0, n__zeros))
with rule ACTIVATE(n__adx(n__zeros)) → ADX(cons(0, n__zeros)) at position [] and matcher [ ]

ADX(cons(0, n__zeros))INCR(cons(0, n__adx(activate(n__zeros))))
with rule ADX(cons(X', L')) → INCR(cons(X', n__adx(activate(L')))) at position [] and matcher [X' / 0, L' / n__zeros]

INCR(cons(0, n__adx(activate(n__zeros))))ACTIVATE(n__adx(activate(n__zeros)))
with rule INCR(cons(X, L)) → ACTIVATE(L)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(32) NO