Termination w.r.t. Q proof of /tmp/tmppmB3NJ/ExIntrod_GM01

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

0 QTRS

↳1 QTRSRRRProof (⇔)

↳2 QTRS

↳3 QTRSRRRProof (⇔)

↳4 QTRS

↳5 QTRSRRRProof (⇔)

↳6 QTRS

↳7 DependencyPairsProof (⇔)

↳8 QDP

↳9 DependencyGraphProof (⇔)

↳10 QDP

↳11 MRRProof (⇔)

↳12 QDP

↳13 QDPOrderProof (⇔)

↳14 QDP

↳15 Narrowing (⇔)

↳16 QDP

↳17 Narrowing (⇔)

↳18 QDP

↳19 Narrowing (⇔)

↳20 QDP

↳21 DependencyGraphProof (⇔)

↳22 QDP

↳23 Narrowing (⇔)

↳24 QDP

↳25 DependencyGraphProof (⇔)

↳26 QDP

↳27 QDPOrderProof (⇔)

↳28 QDP

↳29 QDPOrderProof (⇔)

↳30 QDP

↳31 NonTerminationProof (⇔)

↳32 NO

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

(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) Narrowing (EQUIVALENT transformation)

(16) Obligation:

(17) Narrowing (EQUIVALENT transformation)

(18) Obligation:

(19) Narrowing (EQUIVALENT transformation)

(20) Obligation:

(21) DependencyGraphProof (EQUIVALENT transformation)

(22) Obligation:

(23) Narrowing (EQUIVALENT transformation)

(24) Obligation:

(25) DependencyGraphProof (EQUIVALENT transformation)

(26) Obligation:

(27) QDPOrderProof (EQUIVALENT transformation)

(28) Obligation:

(29) QDPOrderProof (EQUIVALENT transformation)

(30) Obligation:

(31) NonTerminationProof (EQUIVALENT transformation)

(32) NO