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:

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(X)

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:

A__FST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(fst2(X1, X2)) -> MARK1(X2)
MARK1(fst2(X1, X2)) -> MARK1(X1)
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(len1(X)) -> MARK1(X)
A__ADD2(0, X) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
MARK1(len1(X)) -> A__LEN1(mark1(X))
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(fst2(X1, X2)) -> A__FST2(mark1(X1), mark1(X2))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(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:

A__FST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(fst2(X1, X2)) -> MARK1(X2)
MARK1(fst2(X1, X2)) -> MARK1(X1)
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(len1(X)) -> MARK1(X)
A__ADD2(0, X) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
MARK1(len1(X)) -> A__LEN1(mark1(X))
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(fst2(X1, X2)) -> A__FST2(mark1(X1), mark1(X2))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(X)

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

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

A__FST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(fst2(X1, X2)) -> MARK1(X2)
MARK1(fst2(X1, X2)) -> MARK1(X1)
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(len1(X)) -> MARK1(X)
A__ADD2(0, X) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(fst2(X1, X2)) -> A__FST2(mark1(X1), mark1(X2))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
A__FROM1(X) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(X)

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.


A__FST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(fst2(X1, X2)) -> MARK1(X2)
MARK1(fst2(X1, X2)) -> MARK1(X1)
MARK1(add2(X1, X2)) -> MARK1(X2)
MARK1(len1(X)) -> MARK1(X)
A__ADD2(0, X) -> MARK1(X)
MARK1(add2(X1, X2)) -> MARK1(X1)
MARK1(fst2(X1, X2)) -> A__FST2(mark1(X1), mark1(X2))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
The remaining pairs can at least by weakly be oriented.

MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__FROM1(X) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Combined order from the following AFS and order.
A__FST2(x1, x2)  =  A__FST2(x1, x2)
s1(x1)  =  s
cons2(x1, x2)  =  x1
MARK1(x1)  =  x1
fst2(x1, x2)  =  fst2(x1, x2)
add2(x1, x2)  =  add2(x1, x2)
len1(x1)  =  len1(x1)
A__ADD2(x1, x2)  =  A__ADD1(x2)
0  =  0
from1(x1)  =  x1
A__FROM1(x1)  =  x1
mark1(x1)  =  x1
a__from1(x1)  =  x1
a__fst2(x1, x2)  =  a__fst2(x1, x2)
a__len1(x1)  =  a__len1(x1)
nil  =  nil
a__add2(x1, x2)  =  a__add2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[s, fst2, len1, afst2, alen1, nil] > AFST2
[s, fst2, len1, afst2, alen1, nil] > [add2, aadd2] > AADD1
[s, fst2, len1, afst2, alen1, nil] > 0


The following usable rules [14] were oriented:

mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
a__add2(0, X) -> mark1(X)
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
a__len1(X) -> len1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__add2(X1, X2) -> add2(X1, X2)
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__from1(X) -> from1(X)
a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__fst2(X1, X2) -> fst2(X1, X2)



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

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

MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(X)

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(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
A__FROM1(X) -> MARK1(X)
The remaining pairs can at least by weakly be oriented.

MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
from1(x1)  =  from1(x1)
A__FROM1(x1)  =  A__FROM1(x1)
mark1(x1)  =  x1
cons2(x1, x2)  =  x1
a__fst2(x1, x2)  =  x2
s1(x1)  =  s
fst2(x1, x2)  =  x2
a__from1(x1)  =  a__from1(x1)
len1(x1)  =  len1(x1)
a__len1(x1)  =  a__len1(x1)
0  =  0
nil  =  nil
a__add2(x1, x2)  =  a__add1(x2)
add2(x1, x2)  =  add1(x2)

Lexicographic Path Order [19].
Precedence:
[from1, afrom1] > AFROM1
[len1, alen1] > 0
[aadd1, add1]


The following usable rules [14] were oriented:

mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
a__add2(0, X) -> mark1(X)
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
a__len1(X) -> len1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__add2(X1, X2) -> add2(X1, X2)
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__from1(X) -> from1(X)
a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__fst2(X1, X2) -> fst2(X1, X2)



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

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

MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(X)

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(cons2(X1, X2)) -> MARK1(X1)
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)
cons2(x1, x2)  =  cons2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[MARK1, cons2]


The following usable rules [14] were oriented: none



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

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

a__fst2(0, Z) -> nil
a__fst2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), fst2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__add2(0, X) -> mark1(X)
a__add2(s1(X), Y) -> s1(add2(X, Y))
a__len1(nil) -> 0
a__len1(cons2(X, Z)) -> s1(len1(Z))
mark1(fst2(X1, X2)) -> a__fst2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(add2(X1, X2)) -> a__add2(mark1(X1), mark1(X2))
mark1(len1(X)) -> a__len1(mark1(X))
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
mark1(nil) -> nil
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__fst2(X1, X2) -> fst2(X1, X2)
a__from1(X) -> from1(X)
a__add2(X1, X2) -> add2(X1, X2)
a__len1(X) -> len1(X)

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.