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

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 QDPOrderProof (⇔)

↳8 QDP

(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) DependencyPairsProof (EQUIVALENT transformation)

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

(2) 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)
NATS → ADX(zeros)
NATS → ZEROS
TAIL(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(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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(4) 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(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.
We have to consider all minimal (P,Q,R)-chains.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADX(cons(X, L)) → ACTIVATE(L)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__incr(x1) = x1
INCR(x1) = x1
cons(x1, x2) = x2
n__adx(x1) = n__adx(x1)
ADX(x1) = ADX(x1)
activate(x1) = x1
adx(x1) = adx(x1)
incr(x1) = x1
s(x1) = s
nil = nil
n__zeros = n__zeros
zeros = zeros
0 = 0

Lexicographic path order with status [LPO].
Quasi-Precedence:

[n_{_adx₁}, ADX₁, adx₁]
[n_{_zeros}, zeros] > 0

Status:

ADX₁: [1]
n_{_adx₁}: [1]
adx₁: [1]
zeros: []
n_{_zeros}: []
s: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented:

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

(6) 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(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.
We have to consider all minimal (P,Q,R)-chains.

(7) 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: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__incr(x1) = n__incr(x1)
INCR(x1) = x1
cons(x1, x2) = x2
n__adx(x1) = n__adx
ADX(x1) = ADX
activate(x1) = activate
adx(x1) = adx
incr(x1) = x1
s(x1) = x1
nil = nil
n__zeros = n__zeros
zeros = zeros
0 = 0

Lexicographic path order with status [LPO].
Quasi-Precedence:

adx > [n_{_adx}, ADX] > activate
zeros > n_{_zeros}
zeros > 0

Status:

n_{_incr₁}: [1]
ADX: []
n_{_adx}: []
zeros: []
adx: []
n_{_zeros}: []
activate: []
0: []
nil: []

The following usable rules [FROCOS05] were oriented: none

(8) 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(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.
We have to consider all minimal (P,Q,R)-chains.