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__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(zeros) -> A__ZEROS
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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:

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(zeros) -> A__ZEROS
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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:

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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.


A__TAIL1(cons2(X, XS)) -> MARK1(XS)
MARK1(tail1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( zeros ) = 1


POL( mark1(x1) ) = x1


POL( a__zeros ) = 1


POL( MARK1(x1) ) = x1


POL( A__TAIL1(x1) ) = x1 + 1


POL( 0 ) = max{0, -1}


POL( a__tail1(x1) ) = x1 + 1


POL( tail1(x1) ) = x1 + 1


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



The following usable rules [14] were oriented:

a__zeros -> zeros
a__zeros -> cons2(0, zeros)
mark1(zeros) -> a__zeros
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
a__tail1(X) -> tail1(X)
mark1(0) -> 0
mark1(tail1(X)) -> a__tail1(mark1(X))
a__tail1(cons2(X, XS)) -> mark1(XS)



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

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

MARK1(tail1(X)) -> A__TAIL1(mark1(X))
MARK1(cons2(X1, X2)) -> MARK1(X1)

The TRS R consists of the following rules:

a__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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
            ↳ 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__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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 Order [17,21] with Interpretation:

POL( MARK1(x1) ) = x1


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



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__zeros -> cons2(0, zeros)
a__tail1(cons2(X, XS)) -> mark1(XS)
mark1(zeros) -> a__zeros
mark1(tail1(X)) -> a__tail1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(mark1(X1), X2)
mark1(0) -> 0
a__zeros -> zeros
a__tail1(X) -> tail1(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.