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__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(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:

MARK1(first2(X1, X2)) -> MARK1(X1)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(s1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

MARK1(first2(X1, X2)) -> MARK1(X1)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(s1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FROM1(X) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(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(first2(X1, X2)) -> MARK1(X1)
MARK1(first2(X1, X2)) -> A__FIRST2(mark1(X1), mark1(X2))
MARK1(s1(X)) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X2)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
The remaining pairs can at least by weakly be oriented.

A__FROM1(X) -> MARK1(X)
Used ordering: Combined order from the following AFS and order.
MARK1(x1)  =  MARK1(x1)
first2(x1, x2)  =  first2(x1, x2)
A__FIRST2(x1, x2)  =  A__FIRST1(x2)
mark1(x1)  =  x1
s1(x1)  =  s1(x1)
from1(x1)  =  from1(x1)
A__FROM1(x1)  =  A__FROM1(x1)
cons2(x1, x2)  =  cons1(x1)
a__first2(x1, x2)  =  a__first2(x1, x2)
a__from1(x1)  =  a__from1(x1)
nil  =  nil
0  =  0

Lexicographic Path Order [19].
Precedence:
[MARK1, AFIRST1, AFROM1]
[from1, cons1, afrom1] > [first2, afirst2, nil]
[from1, cons1, afrom1] > s1
0 > [first2, afirst2, nil]


The following usable rules [14] were oriented:

mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
a__from1(X) -> from1(X)
a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__first2(X1, X2) -> first2(X1, X2)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

A__FROM1(X) -> MARK1(X)

The TRS R consists of the following rules:

a__first2(0, X) -> nil
a__first2(s1(X), cons2(Y, Z)) -> cons2(mark1(Y), first2(X, Z))
a__from1(X) -> cons2(mark1(X), from1(s1(X)))
mark1(first2(X1, X2)) -> a__first2(mark1(X1), mark1(X2))
mark1(from1(X)) -> a__from1(mark1(X))
mark1(0) -> 0
mark1(nil) -> nil
mark1(s1(X)) -> s1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__first2(X1, X2) -> first2(X1, X2)
a__from1(X) -> from1(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.