Termination w.r.t. Q proof of TRS/TRCSREx4_7_77_Bor03

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.

MARK₁(cons₂(X1, X2)) -> MARK₁(X1)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__TAIL₁(x₁)) = x₁
POL(MARK₁(x₁)) = 1 + x₁
POL(a__tail₁(x₁)) = x₁
POL(a__zeros) = 1
POL(cons₂(x₁, x₂)) = 1 + x₁ + x₂
POL(mark₁(x₁)) = 1 + x₁
POL(tail₁(x₁)) = x₁
POL(zeros) = 0

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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)

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(tail₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

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

Used ordering: Polynomial interpretation [21]:

POL(0) = 1
POL(A__TAIL₁(x₁)) = x₁
POL(MARK₁(x₁)) = x₁
POL(a__tail₁(x₁)) = 1 + x₁
POL(a__zeros) = 1
POL(cons₂(x₁, x₂)) = x₁ + x₂
POL(mark₁(x₁)) = 1 + x₁
POL(tail₁(x₁)) = 1 + x₁
POL(zeros) = 0

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ 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)

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(tail₁(X)) -> A__TAIL₁(mark₁(X))

The remaining pairs can at least be oriented weakly.

A__TAIL₁(cons₂(X, XS)) -> MARK₁(XS)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__TAIL₁(x₁)) = x₁
POL(MARK₁(x₁)) = x₁
POL(a__tail₁(x₁)) = 1 + x₁
POL(a__zeros) = 0
POL(cons₂(x₁, x₂)) = x₂
POL(mark₁(x₁)) = x₁
POL(tail₁(x₁)) = 1 + x₁
POL(zeros) = 0

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

A__TAIL₁(cons₂(X, XS)) -> MARK₁(XS)

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 0 SCCs with 1 less node.