Termination w.r.t. Q proof of /tmp/tmpowTQ8w/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 DependencyPairsProof (⇔)

↳6 QDP

↳7 DependencyGraphProof (⇔)

↳8 QDP

↳9 MRRProof (⇔)

↳10 QDP

↳11 QDPOrderProof (⇔)

↳12 QDP

↳13 Instantiation (⇔)

↳14 QDP

↳15 NonTerminationProof (⇔)

↳16 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(X)
activate(n__adx(X)) → adx(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 + x₁
POL(incr(x₁)) = x₁
POL(n__adx(x₁)) = 2·x₁
POL(n__incr(x₁)) = x₁
POL(n__zeros) = 0
POL(nats) = 1
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:

nats → adx(zeros)
head(cons(X, L)) → X
tail(cons(X, L)) → activate(L)

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

Q is empty.

(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(incr(x₁)) = x₁
POL(n__adx(x₁)) = 2·x₁
POL(n__incr(x₁)) = x₁
POL(n__zeros) = 0
POL(nil) = 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:

adx(nil) → nil

(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)
incr(X) → n__incr(X)
adx(X) → n__adx(X)
zeros → n__zeros
activate(n__incr(X)) → incr(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) 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(X)
ACTIVATE(n__adx(X)) → ADX(X)
ACTIVATE(n__zeros) → ZEROS

The TRS R consists of the following rules:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(9) 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)

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__adx(x₁)) = 1 + x₁
POL(n__incr(x₁)) = x₁
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x₁)) = x₁
POL(zeros) = 0

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) 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 + 1

POL( n__adx(x₁) ) = 0

POL( activate(x₁) ) = 0

POL( n__incr(x₁) ) = x₁

POL( incr(x₁) ) = max{0, 2x₁ - 1}

POL( adx(x₁) ) = 2x₁ + 1

POL( n__zeros ) = 1

POL( zeros ) = 1

POL( s(x₁) ) = x₁ + 2

POL( nil ) = 2

POL( 0 ) = 2

POL( ACTIVATE(x₁) ) = x₁ + 2

POL( ADX(x₁) ) = 2

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(13) Instantiation (EQUIVALENT transformation)

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

INCR(cons(y_0, n__adx(y_2))) → ACTIVATE(n__adx(y_2))

(14) Obligation:

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

ACTIVATE(n__adx(X)) → ADX(X)
ADX(cons(X, L)) → INCR(cons(X, n__adx(activate(L))))
INCR(cons(y_0, n__adx(y_2))) → ACTIVATE(n__adx(y_2))

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(X)
activate(n__adx(X)) → adx(X)
activate(n__zeros) → zeros
activate(X) → X

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

(15) 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(0, n__zeros))
with rule zeros → cons(0, n__zeros) at position [0] 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(y_0, n__adx(y_2))) → ACTIVATE(n__adx(y_2)) at position [] and matcher [y_0 / 0, y_2 / 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.

(0) Obligation:

(1) QTRSRRRProof (EQUIVALENT transformation)

(2) Obligation:

(3) QTRSRRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) MRRProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) Instantiation (EQUIVALENT transformation)

(14) Obligation:

(15) NonTerminationProof (EQUIVALENT transformation)

(16) NO