Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

INCR(cons(X, Y)) → ACTIVATE(Y)
ADX(cons(X, Y)) → ACTIVATE(X)
ADX(cons(X, Y)) → ACTIVATE(Y)
NATS → ZEROS
ACTIVATE(n__incr(X)) → INCR(activate(X))
NATS → ADX(zeros)
ACTIVATE(n__s(X)) → S(X)
HD(cons(X, Y)) → ACTIVATE(X)
TL(cons(X, Y)) → ACTIVATE(Y)
INCR(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ACTIVATE(n__0) → 01
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ACTIVATE(X)
ACTIVATE(n__zeros) → ZEROS

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

INCR(cons(X, Y)) → ACTIVATE(Y)
ADX(cons(X, Y)) → ACTIVATE(X)
ADX(cons(X, Y)) → ACTIVATE(Y)
NATS → ZEROS
ACTIVATE(n__incr(X)) → INCR(activate(X))
NATS → ADX(zeros)
ACTIVATE(n__s(X)) → S(X)
HD(cons(X, Y)) → ACTIVATE(X)
TL(cons(X, Y)) → ACTIVATE(Y)
INCR(cons(X, Y)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ACTIVATE(n__0) → 01
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
ACTIVATE(n__adx(X)) → ACTIVATE(X)
ACTIVATE(n__zeros) → ZEROS

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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 7 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

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.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ADX(cons(X, Y)) → ACTIVATE(X)
ADX(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__adx(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.

INCR(cons(X, Y)) → ACTIVATE(Y)
ACTIVATE(n__adx(X)) → ADX(activate(X))
ADX(cons(X, Y)) → INCR(cons(activate(X), n__adx(activate(Y))))
ACTIVATE(n__incr(X)) → INCR(activate(X))
ACTIVATE(n__incr(X)) → ACTIVATE(X)
INCR(cons(X, Y)) → ACTIVATE(X)
Used ordering: Polynomial interpretation [25,35]:

POL(n__incr(x1)) = x_1   
POL(ADX(x1)) = 1 + x_1   
POL(n__adx(x1)) = 1 + x_1   
POL(n__zeros) = 0   
POL(activate(x1)) = x_1   
POL(n__s(x1)) = x_1   
POL(0) = 0   
POL(cons(x1, x2)) = (4)x_1 + x_2   
POL(adx(x1)) = 1 + x_1   
POL(incr(x1)) = x_1   
POL(zeros) = 0   
POL(n__0) = 0   
POL(s(x1)) = x_1   
POL(INCR(x1)) = x_1   
POL(ACTIVATE(x1)) = x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP

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

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

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