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 oriented strictly 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 be oriented weakly.

A__FROM1(X) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(0) = 1   
POL(A__FIRST2(x1, x2)) = 3 + 3·x1 + 3·x1·x2 + 3·x2   
POL(A__FROM1(x1)) = 3 + 3·x1   
POL(MARK1(x1)) = 3 + 3·x1   
POL(a__first2(x1, x2)) = 3 + x1 + x1·x2 + 2·x2   
POL(a__from1(x1)) = 3 + 2·x1   
POL(cons2(x1, x2)) = 3 + 2·x1   
POL(first2(x1, x2)) = 3 + x1 + x1·x2 + 2·x2   
POL(from1(x1)) = 3 + 2·x1   
POL(mark1(x1)) = x1   
POL(nil) = 3   
POL(s1(x1)) = 3 + 2·x1 + x12   

The following usable rules [14] were oriented:

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