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(first2(X1, X2)) -> MARK1(X2)
MARK1(from1(X)) -> MARK1(X)
MARK1(from1(X)) -> A__FROM1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
The remaining pairs can at least be oriented weakly.

MARK1(s1(X)) -> MARK1(X)
A__FROM1(X) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK1(x1) ) = x1


POL( first2(x1, x2) ) = x1 + x2 + 1


POL( A__FIRST2(x1, x2) ) = max{0, x2 - 1}


POL( mark1(x1) ) = x1


POL( s1(x1) ) = x1


POL( from1(x1) ) = x1 + 1


POL( A__FROM1(x1) ) = x1


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


POL( nil ) = 0


POL( a__first2(x1, x2) ) = x1 + x2 + 1


POL( 0 ) = 0


POL( a__from1(x1) ) = x1 + 1



The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ DependencyGraphProof

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

MARK1(s1(X)) -> MARK1(X)
A__FIRST2(s1(X), cons2(Y, Z)) -> MARK1(Y)
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 1 SCC with 2 less nodes.

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

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

MARK1(s1(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MARK1(s1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MARK1(x1) ) = x1


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.