Termination w.r.t. Q of the given QTRS could be disproven:

(0) Obligation:

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
tl(cons(X, Y)) → activate(Y)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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)) = 2·x₁ + x₂   
POL(hd(x₁)) = 2·x₁   
POL(incr(x₁)) = x₁   
POL(n__0) = 0   
POL(n__adx(x₁)) = 2·x₁   
POL(n__incr(x₁)) = x₁   
POL(n__s(x₁)) = x₁   
POL(n__zeros) = 0   
POL(nats) = 0   
POL(s(x₁)) = x₁   
POL(tl(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:

tl(cons(X, Y)) → activate(Y)

(2) Obligation:

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
hd(cons(X, Y)) → activate(X)
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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)) = 2·x₁ + 2·x₂   
POL(hd(x₁)) = 2 + 2·x₁   
POL(incr(x₁)) = x₁   
POL(n__0) = 0   
POL(n__adx(x₁)) = 2·x₁   
POL(n__incr(x₁)) = x₁   
POL(n__s(x₁)) = x₁   
POL(n__zeros) = 0   
POL(nats) = 0   
POL(s(x₁)) = x₁   
POL(zeros) = 0   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

hd(cons(X, Y)) → activate(X)

(4) Obligation:

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

nats → adx(zeros)
zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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 + 2·x₂   
POL(incr(x₁)) = x₁   
POL(n__0) = 0   
POL(n__adx(x₁)) = x₁   
POL(n__incr(x₁)) = x₁   
POL(n__s(x₁)) = x₁   
POL(n__zeros) = 0   
POL(nats) = 2   
POL(s(x₁)) = x₁   
POL(zeros) = 0   

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

nats → adx(zeros)

(6) Obligation:

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

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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, Y)) → ACTIVATE(X)
INCR(cons(X, Y)) → ACTIVATE(Y)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ADX(cons(X, Y)) → ACTIVATE(X)
ADX(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__0) → 0¹
ACTIVATE(n__zeros) → ZEROS
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__incr(X)) → INCR(X)
ACTIVATE(n__adx(X)) → ADX(X)

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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 3 less nodes.

(10) Obligation:

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

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

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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, Y)) → ACTIVATE(X)
ADX(cons(X, Y)) → ACTIVATE(Y)

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x₁)) = 2·x₁   
POL(ADX(x₁)) = 2 + 2·x₁   
POL(INCR(x₁)) = 2·x₁   
POL(activate(x₁)) = x₁   
POL(adx(x₁)) = 1 + x₁   
POL(cons(x₁, x₂)) = 2·x₁ + x₂   
POL(incr(x₁)) = x₁   
POL(n__0) = 0   
POL(n__adx(x₁)) = 1 + x₁   
POL(n__incr(x₁)) = x₁   
POL(n__s(x₁)) = x₁   
POL(n__zeros) = 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(X)
INCR(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
INCR(cons(X, Y)) → ACTIVATE(Y)

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
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.

INCR(cons(X, Y)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(ACTIVATE(x1)) =

-I

+

0A

·

x1

POL(n__incr(x1)) =

-I

+

0A

·

x1

POL(INCR(x1)) =

-I

+

0A

·

x1

POL(cons(x1, x2)) =

-I

+

1A

·

x1

+

0A

·

x2

POL(n__adx(x1)) =

-I

+

0A

·

x1

POL(ADX(x1)) =

-I

+

0A

·

x1

POL(activate(x1)) =

-I

+

0A

·

x1

POL(n__0) =

0A

POL(0) =

0A

POL(n__zeros) =

1A

POL(zeros) =

1A

POL(n__s(x1)) =

-I

+

0A

·

x1

POL(s(x1)) =

-I

+

0A

·

x1

POL(incr(x1)) =

-I

+

0A

·

x1

POL(adx(x1)) =

-I

+

0A

·

x1

The following usable rules [FROCOS05] were oriented:

activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X
incr(X) → n__incr(X)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(X) → n__adx(X)
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
zeros → cons(n__0, n__zeros)
zeros → n__zeros
0 → n__0
s(X) → n__s(X)

(14) Obligation:

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

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

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

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

(15) 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: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( INCR(x₁) ) = x₁ + 1

POL( cons(x₁, x₂) ) = 2x₂

POL( n__adx(x₁) ) = max{0, -2}

POL( activate(x₁) ) = 1

POL( n__0 ) = 1

POL( 0 ) = 2

POL( n__zeros ) = 0

POL( zeros ) = 2

POL( n__s(x₁) ) = x₁ + 2

POL( s(x₁) ) = max{0, 2x₁ - 2}

POL( n__incr(x₁) ) = 2x₁ + 2

POL( incr(x₁) ) = max{0, x₁ - 2}

POL( adx(x₁) ) = x₁ + 2

POL( ACTIVATE(x₁) ) = 2x₁ + 1

POL( ADX(x₁) ) = 1

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

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

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

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

(17) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule INCR(cons(X, Y)) → ACTIVATE(Y) we obtained the following new rules [LPAR04]:

INCR(cons(y_1, n__adx(y_3))) → ACTIVATE(n__adx(y_3))

(18) Obligation:

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

ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
INCR(cons(y_1, n__adx(y_3))) → ACTIVATE(n__adx(y_3))

The TRS R consists of the following rules:

zeros → cons(n__0, n__zeros)
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(activate(Y)))
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
0 → n__0
zeros → n__zeros
s(X) → n__s(X)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(X) → X

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

(19) 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 = ADX(activate(n__zeros)) evaluates to t =ADX(activate(n__zeros))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

ADX(activate(n__zeros)) → ADX(zeros)
with rule activate(n__zeros) → zeros at position [0] and matcher [ ]

ADX(zeros) → ADX(cons(n__0, n__zeros))
with rule zeros → cons(n__0, n__zeros) at position [0] and matcher [ ]

ADX(cons(n__0, n__zeros)) → INCR(cons(activate(n__0), n__adx(activate(n__zeros))))
with rule ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y)))) at position [] and matcher [X / n__0, Y / n__zeros]

INCR(cons(activate(n__0), n__adx(activate(n__zeros)))) → ACTIVATE(n__adx(activate(n__zeros)))
with rule INCR(cons(y_1, n__adx(y_3))) → ACTIVATE(n__adx(y_3)) at position [] and matcher [y_1 / activate(n__0), y_3 / activate(n__zeros)]

ACTIVATE(n__adx(activate(n__zeros))) → ADX(activate(n__zeros))
with rule ACTIVATE(n__adx(X)) → ADX(X)

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.

(20) NO