Termination w.r.t. Q proof of Transformed_CSR_04_ExIntrod_GM04

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

0 QTRS

↳1 QTRSRRRProof (⇔, 0 ms)

↳2 QTRS

↳3 QTRSRRRProof (⇔, 1 ms)

↳4 QTRS

↳5 QTRSRRRProof (⇔, 0 ms)

↳6 QTRS

↳7 DependencyPairsProof (⇔, 0 ms)

↳8 QDP

↳9 DependencyGraphProof (⇔, 0 ms)

↳10 QDP

↳11 MRRProof (⇔, 0 ms)

↳12 QDP

↳13 QDPOrderProof (⇔, 5 ms)

↳14 QDP

↳15 QDPOrderProof (⇔, 0 ms)

↳16 QDP

↳17 TransformationProof (⇔, 0 ms)

↳18 QDP

↳19 TransformationProof (⇔, 0 ms)

↳20 QDP

↳21 TransformationProof (⇔, 0 ms)

↳22 QDP

↳23 DependencyGraphProof (⇔, 0 ms)

↳24 QDP

↳25 TransformationProof (⇔, 0 ms)

↳26 QDP

↳27 DependencyGraphProof (⇔, 0 ms)

↳28 QDP

↳29 TransformationProof (⇔, 0 ms)

↳30 QDP

↳31 DependencyGraphProof (⇔, 0 ms)

↳32 QDP

↳33 TransformationProof (⇔, 0 ms)

↳34 QDP

↳35 DependencyGraphProof (⇔, 0 ms)

↳36 QDP

↳37 TransformationProof (⇔, 0 ms)

↳38 QDP

↳39 DependencyGraphProof (⇔, 0 ms)

↳40 QDP

↳41 TransformationProof (⇔, 0 ms)

↳42 QDP

↳43 DependencyGraphProof (⇔, 0 ms)

↳44 QDP

↳45 QDPOrderProof (⇔, 13 ms)

↳46 QDP

↳47 QDPOrderProof (⇔, 51 ms)

↳48 QDP

↳49 NonTerminationLoopProof (⇒, 0 ms)

↳50 NO

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

hd(cons(X, Y)) → activate(X)

(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))))
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(activate(X))
activate(n__adx(X)) → adx(activate(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)) = 1 + x₁
POL(cons(x₁, x₂)) = 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(nats) = 2
POL(s(x₁)) = x₁
POL(tl(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)

(4) 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))))
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(activate(X))
activate(n__adx(X)) → adx(activate(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)) = 2 + 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = x₁
POL(n__0) = 0
POL(n__adx(x₁)) = 2 + 2·x₁
POL(n__incr(x₁)) = x₁
POL(n__s(x₁)) = x₁
POL(n__zeros) = 0
POL(s(x₁)) = x₁
POL(tl(x₁)) = 2 + 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)

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

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

Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0
POL(ACTIVATE(x₁)) = x₁
POL(ADX(x₁)) = 2 + 2·x₁
POL(INCR(x₁)) = x₁
POL(activate(x₁)) = x₁
POL(adx(x₁)) = 2 + 2·x₁
POL(cons(x₁, x₂)) = 2·x₁ + x₂
POL(incr(x₁)) = x₁
POL(n__0) = 0
POL(n__adx(x₁)) = 2 + 2·x₁
POL(n__incr(x₁)) = x₁
POL(n__s(x₁)) = x₁
POL(n__zeros) = 2
POL(s(x₁)) = x₁
POL(zeros) = 2

(12) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(activate(X))
INCR(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ADX(activate(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(activate(X))
activate(n__adx(X)) → adx(activate(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,JAR06].

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)) = 5A + 5A · x₁

POL(n__incr(x₁)) = 5A + 0A · x₁

POL(INCR(x₁)) = 5A + 5A · x₁

POL(activate(x₁)) = 0A + 0A · x₁

POL(cons(x₁, x₂)) = 5A + 1A · x₁ + 0A · x₂

POL(n__adx(x₁)) = 1A + 0A · x₁

POL(ADX(x₁)) = 5A + 5A · x₁

POL(n__0) = 0A

POL(0) = 0A

POL(n__zeros) = 5A

POL(zeros) = 5A

POL(n__s(x₁)) = -I + 0A · x₁

POL(s(x₁)) = 0A + 0A · x₁

POL(incr(x₁)) = 5A + 0A · x₁

POL(adx(x₁)) = 1A + 0A · x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

activate(n__0) → 0
activate(n__zeros) → zeros
activate(n__s(X)) → s(X)
activate(n__incr(X)) → incr(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X
incr(X) → n__incr(X)
adx(X) → n__adx(X)
adx(cons(X, Y)) → incr(cons(activate(X), n__adx(activate(Y))))
incr(cons(X, Y)) → cons(n__s(activate(X)), n__incr(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(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ADX(activate(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(activate(X))
activate(n__adx(X)) → adx(activate(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,JAR06].

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) ) = 2

POL( INCR(x₁) ) = x₁ + 1

POL( activate(x₁) ) = x₁

POL( n__0 ) = 1

POL( 0 ) = 1

POL( n__zeros ) = 2

POL( zeros ) = 2

POL( n__s(x₁) ) = 0

POL( s(x₁) ) = 0

POL( n__incr(x₁) ) = x₁ + 1

POL( incr(x₁) ) = x₁ + 1

POL( n__adx(x₁) ) = 2

POL( adx(x₁) ) = 2

POL( cons(x₁, x₂) ) = max{0, x₂ - 1}

POL( ACTIVATE(x₁) ) = x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(16) Obligation:

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

ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__adx(X)) → ADX(activate(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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(17) TransformationProof (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__0)) → INCR(0) → ACTIVATE(n__incr(n__0)) → INCR(0)
ACTIVATE(n__incr(n__zeros)) → INCR(zeros) → ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0)) → ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0))
ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0))) → ACTIVATE(n__incr(n__incr(x0))) → INCR(incr(activate(x0)))
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0))) → ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0) → ACTIVATE(n__incr(x0)) → INCR(x0)

(18) Obligation:

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

ACTIVATE(n__adx(X)) → ADX(activate(X))
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__0)) → INCR(0)
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0))
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)

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(19) TransformationProof (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__0)) → ADX(0) → ACTIVATE(n__adx(n__0)) → ADX(0)
ACTIVATE(n__adx(n__zeros)) → ADX(zeros) → ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0)) → ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(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__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0) → ACTIVATE(n__adx(x0)) → ADX(x0)

(20) Obligation:

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

ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__0)) → INCR(0)
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0))
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__0)) → ADX(0)
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(21) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__incr(n__0)) → INCR(0) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__incr(n__0)) → INCR(n__0) → ACTIVATE(n__incr(n__0)) → INCR(n__0)

(22) Obligation:

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

ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0))
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__0)) → ADX(0)
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__0)) → INCR(n__0)

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__zeros)) → INCR(zeros)
ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0))
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__0)) → ADX(0)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(25) TransformationProof (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(n__0, n__zeros)) → ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__incr(n__zeros)) → INCR(n__zeros) → ACTIVATE(n__incr(n__zeros)) → INCR(n__zeros)

(26) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0))
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__0)) → ADX(0)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__incr(n__zeros)) → INCR(n__zeros)

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(27) DependencyGraphProof (EQUIVALENT transformation)

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

(28) Obligation:

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

ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0))
INCR(cons(X, Y)) → ACTIVATE(Y)
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__0)) → ADX(0)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(29) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__incr(n__s(x0))) → INCR(s(x0)) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__incr(n__s(x0))) → INCR(n__s(x0)) → ACTIVATE(n__incr(n__s(x0))) → INCR(n__s(x0))

(30) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
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__0)) → ADX(0)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__incr(n__s(x0))) → INCR(n__s(x0))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(31) DependencyGraphProof (EQUIVALENT transformation)

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

(32) 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, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__0)) → ADX(0)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(33) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__adx(n__0)) → ADX(0) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__adx(n__0)) → ADX(n__0) → ACTIVATE(n__adx(n__0)) → ADX(n__0)

(34) 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, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__zeros)) → ADX(zeros)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__0)) → ADX(n__0)

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
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__zeros)) → ADX(zeros)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(37) TransformationProof (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(n__0, n__zeros)) → ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(n__zeros) → ACTIVATE(n__adx(n__zeros)) → ADX(n__zeros)

(38) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
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)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(n__zeros)

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(39) DependencyGraphProof (EQUIVALENT transformation)

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

(40) 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, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0))
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(41) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule ACTIVATE(n__adx(n__s(x0))) → ADX(s(x0)) at position [0] we obtained the following new rules [LPAR04]:

ACTIVATE(n__adx(n__s(x0))) → ADX(n__s(x0)) → ACTIVATE(n__adx(n__s(x0))) → ADX(n__s(x0))

(42) 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, Y)) → ACTIVATE(Y)
ACTIVATE(n__incr(n__adx(x0))) → INCR(adx(activate(x0)))
ACTIVATE(n__incr(x0)) → INCR(x0)
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__incr(x0))) → ADX(incr(activate(x0)))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__s(x0))) → ADX(n__s(x0))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(43) DependencyGraphProof (EQUIVALENT transformation)

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

(44) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
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, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__0, n__zeros))
ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

ACTIVATE(n__incr(n__zeros)) → INCR(cons(n__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( adx(x₁) ) = max{0, -2}

POL( incr(x₁) ) = 2x₁

POL( INCR(x₁) ) = x₁ + 2

POL( cons(x₁, x₂) ) = x₂

POL( n__s(x₁) ) = 1

POL( n__incr(x₁) ) = 2x₁

POL( n__adx(x₁) ) = 0

POL( activate(x₁) ) = x₁

POL( n__0 ) = 0

POL( 0 ) = 0

POL( n__zeros ) = 1

POL( zeros ) = 1

POL( s(x₁) ) = 1

POL( ACTIVATE(x₁) ) = x₁ + 2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(46) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
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, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(n__adx(x0))) → ADX(adx(activate(x0)))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

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)) = [4]x₁
POL(ADX(x₁)) = [2] + [2]x₁
POL(INCR(x₁)) = [2]x₁
POL(activate(x₁)) = x₁
POL(adx(x₁)) = [1/2] + x₁
POL(cons(x₁, x₂)) = [4]x₁ + [2]x₂
POL(incr(x₁)) = [1/2]x₁
POL(n__0) = 0
POL(n__adx(x₁)) = [1/2] + x₁
POL(n__incr(x₁)) = [1/2]x₁
POL(n__s(x₁)) = 0
POL(n__zeros) = 0
POL(s(x₁)) = 0
POL(zeros) = 0

The value of delta used in the strict ordering is 1.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(48) Obligation:

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

INCR(cons(X, Y)) → ACTIVATE(Y)
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, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__adx(x0)) → ADX(x0)
ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros))

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(activate(X))
activate(n__adx(X)) → adx(activate(X))
activate(X) → X

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

(49) NonTerminationLoopProof (COMPLETE 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(n__0, n__zeros))
with rule ACTIVATE(n__adx(n__zeros)) → ADX(cons(n__0, n__zeros)) at position [] 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(X, Y)) → ACTIVATE(Y)

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.

(0) Obligation:

(1) QTRSRRRProof (EQUIVALENT transformation)

(2) Obligation:

(3) QTRSRRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) QTRSRRRProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyPairsProof (EQUIVALENT transformation)

(8) Obligation:

(9) DependencyGraphProof (EQUIVALENT transformation)

(10) Obligation:

(11) MRRProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) TransformationProof (EQUIVALENT transformation)

(18) Obligation:

(19) TransformationProof (EQUIVALENT transformation)

(20) Obligation:

(21) TransformationProof (EQUIVALENT transformation)

(22) Obligation:

(23) DependencyGraphProof (EQUIVALENT transformation)

(24) Obligation:

(25) TransformationProof (EQUIVALENT transformation)

(26) Obligation:

(27) DependencyGraphProof (EQUIVALENT transformation)

(28) Obligation:

(29) TransformationProof (EQUIVALENT transformation)

(30) Obligation:

(31) DependencyGraphProof (EQUIVALENT transformation)

(32) Obligation:

(33) TransformationProof (EQUIVALENT transformation)

(34) Obligation:

(35) DependencyGraphProof (EQUIVALENT transformation)

(36) Obligation:

(37) TransformationProof (EQUIVALENT transformation)

(38) Obligation:

(39) DependencyGraphProof (EQUIVALENT transformation)

(40) Obligation:

(41) TransformationProof (EQUIVALENT transformation)

(42) Obligation:

(43) DependencyGraphProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) NonTerminationLoopProof (COMPLETE transformation)

(50) NO