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 oriented strictly and are deleted.


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(fst2(X1, X2)) -> A__FST2(mark1(X1), mark1(X2))
A__FROM1(X) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

A__FST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A__ADD2(x1, x2)) = 1 + x2   
POL(A__FROM1(x1)) = 1 + x1   
POL(A__FST2(x1, x2)) = x2   
POL(MARK1(x1)) = x1   
POL(a__add2(x1, x2)) = 1 + x1 + x2   
POL(a__from1(x1)) = 1 + x1   
POL(a__fst2(x1, x2)) = 1 + x1 + x2   
POL(a__len1(x1)) = 1 + x1   
POL(add2(x1, x2)) = 1 + x1 + x2   
POL(cons2(x1, x2)) = x1   
POL(from1(x1)) = 1 + x1   
POL(fst2(x1, x2)) = 1 + x1 + x2   
POL(len1(x1)) = 1 + x1   
POL(mark1(x1)) = x1   
POL(nil) = 0   
POL(s1(x1)) = 0   

The following usable rules [14] were oriented:

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



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

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

A__FST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(add2(X1, X2)) -> A__ADD2(mark1(X1), mark1(X2))
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
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 oriented strictly and are deleted.


MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x1   
POL(cons2(x1, x2)) = 1 + x1   

The following usable rules [14] were oriented: none



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