Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MARK1(nats) -> NATS_ACTIVE
MARK1(tl1(x)) -> MARK1(x)
ADD_ACTIVE1(cons2(x, y)) -> INCR_ACTIVE1(cons2(x, add1(y)))
MARK1(tl1(x)) -> TL_ACTIVE1(mark1(x))
MARK1(add1(x)) -> MARK1(x)
HD_ACTIVE1(cons2(x, y)) -> MARK1(x)
MARK1(zeros) -> ZEROS_ACTIVE
MARK1(incr1(x)) -> INCR_ACTIVE1(mark1(x))
MARK1(add1(x)) -> ADD_ACTIVE1(mark1(x))
MARK1(hd1(x)) -> HD_ACTIVE1(mark1(x))
NATS_ACTIVE -> ZEROS_ACTIVE
MARK1(hd1(x)) -> MARK1(x)
NATS_ACTIVE -> ADD_ACTIVE1(zeros_active)
TL_ACTIVE1(cons2(x, y)) -> MARK1(y)
MARK1(incr1(x)) -> MARK1(x)

The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

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:

MARK1(nats) -> NATS_ACTIVE
MARK1(tl1(x)) -> MARK1(x)
ADD_ACTIVE1(cons2(x, y)) -> INCR_ACTIVE1(cons2(x, add1(y)))
MARK1(tl1(x)) -> TL_ACTIVE1(mark1(x))
MARK1(add1(x)) -> MARK1(x)
HD_ACTIVE1(cons2(x, y)) -> MARK1(x)
MARK1(zeros) -> ZEROS_ACTIVE
MARK1(incr1(x)) -> INCR_ACTIVE1(mark1(x))
MARK1(add1(x)) -> ADD_ACTIVE1(mark1(x))
MARK1(hd1(x)) -> HD_ACTIVE1(mark1(x))
NATS_ACTIVE -> ZEROS_ACTIVE
MARK1(hd1(x)) -> MARK1(x)
NATS_ACTIVE -> ADD_ACTIVE1(zeros_active)
TL_ACTIVE1(cons2(x, y)) -> MARK1(y)
MARK1(incr1(x)) -> MARK1(x)

The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 7 less nodes.

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

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

MARK1(tl1(x)) -> MARK1(x)
MARK1(tl1(x)) -> TL_ACTIVE1(mark1(x))
MARK1(add1(x)) -> MARK1(x)
HD_ACTIVE1(cons2(x, y)) -> MARK1(x)
TL_ACTIVE1(cons2(x, y)) -> MARK1(y)
MARK1(hd1(x)) -> HD_ACTIVE1(mark1(x))
MARK1(incr1(x)) -> MARK1(x)
MARK1(hd1(x)) -> MARK1(x)

The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MARK1(tl1(x)) -> MARK1(x)
HD_ACTIVE1(cons2(x, y)) -> MARK1(x)
TL_ACTIVE1(cons2(x, y)) -> MARK1(y)
MARK1(hd1(x)) -> HD_ACTIVE1(mark1(x))
MARK1(hd1(x)) -> MARK1(x)
The remaining pairs can at least by weakly be oriented.

MARK1(tl1(x)) -> TL_ACTIVE1(mark1(x))
MARK1(add1(x)) -> MARK1(x)
MARK1(incr1(x)) -> MARK1(x)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
tl1(x1)  =  tl1(x1)
TL_ACTIVE1(x1)  =  TL_ACTIVE1(x1)
mark1(x1)  =  mark1(x1)
add1(x1)  =  x1
HD_ACTIVE1(x1)  =  x1
cons2(x1, x2)  =  cons2(x1, x2)
hd1(x1)  =  hd1(x1)
incr1(x1)  =  x1
hd_active1(x1)  =  hd_active1(x1)
incr_active1(x1)  =  x1
s1(x1)  =  s
nats_active  =  nats_active
nats  =  nats
0  =  0
tl_active1(x1)  =  tl_active1(x1)
add_active1(x1)  =  x1
zeros  =  zeros
zeros_active  =  zeros_active

Lexicographic Path Order [19].
Precedence:
[tl1, mark1, hd1, hdactive1, tlactive1] > zerosactive > 0 > [MARK1, TLACTIVE1, cons2, s]
[tl1, mark1, hd1, hdactive1, tlactive1] > zerosactive > zeros > [MARK1, TLACTIVE1, cons2, s]
[natsactive, nats] > zerosactive > 0 > [MARK1, TLACTIVE1, cons2, s]
[natsactive, nats] > zerosactive > zeros > [MARK1, TLACTIVE1, cons2, s]


The following usable rules [14] were oriented:

mark1(nats) -> nats_active
mark1(zeros) -> zeros_active
mark1(incr1(x)) -> incr_active1(mark1(x))
mark1(add1(x)) -> add_active1(mark1(x))
mark1(hd1(x)) -> hd_active1(mark1(x))
mark1(tl1(x)) -> tl_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
hd_active1(cons2(x, y)) -> mark1(x)
mark1(0) -> 0
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)
hd_active1(x) -> hd1(x)
tl_active1(x) -> tl1(x)
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
add_active1(x) -> add1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
incr_active1(x) -> incr1(x)
zeros_active -> cons2(0, zeros)
zeros_active -> zeros
nats_active -> add_active1(zeros_active)
nats_active -> nats



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

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

MARK1(add1(x)) -> MARK1(x)
MARK1(tl1(x)) -> TL_ACTIVE1(mark1(x))
MARK1(incr1(x)) -> MARK1(x)

The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

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

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

MARK1(add1(x)) -> MARK1(x)
MARK1(incr1(x)) -> MARK1(x)

The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MARK1(incr1(x)) -> MARK1(x)
The remaining pairs can at least by weakly be oriented.

MARK1(add1(x)) -> MARK1(x)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
add1(x1)  =  x1
incr1(x1)  =  incr1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

MARK1(add1(x)) -> MARK1(x)

The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


MARK1(add1(x)) -> MARK1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
add1(x1)  =  add1(x1)

Lexicographic Path Order [19].
Precedence:
add1 > MARK1


The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

nats_active -> add_active1(zeros_active)
hd_active1(x) -> hd1(x)
zeros_active -> cons2(0, zeros)
tl_active1(x) -> tl1(x)
incr_active1(cons2(x, y)) -> cons2(s1(x), incr1(y))
mark1(nats) -> nats_active
add_active1(cons2(x, y)) -> incr_active1(cons2(x, add1(y)))
mark1(zeros) -> zeros_active
hd_active1(cons2(x, y)) -> mark1(x)
mark1(incr1(x)) -> incr_active1(mark1(x))
tl_active1(cons2(x, y)) -> mark1(y)
mark1(add1(x)) -> add_active1(mark1(x))
nats_active -> nats
mark1(hd1(x)) -> hd_active1(mark1(x))
zeros_active -> zeros
mark1(tl1(x)) -> tl_active1(mark1(x))
incr_active1(x) -> incr1(x)
mark1(0) -> 0
add_active1(x) -> add1(x)
mark1(s1(x)) -> s1(x)
mark1(cons2(x, y)) -> cons2(x, y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.