Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/TRCSRSLASHEx4_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 strictly oriented and are deleted.

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

The remaining pairs can at least by weakly be oriented.

MARK₁(tail₁(X)) -> A__TAIL₁(mark₁(X))

Used ordering: Combined order from the following AFS and order.
MARK₁(x1) = MARK₁(x1)
tail₁(x1) = tail₁(x1)
A__TAIL₁(x1) = A__TAIL₁(x1)
mark₁(x1) = mark₁(x1)
cons₂(x1, x2) = cons₂(x1, x2)
zeros = zeros
a__zeros = a__zeros
a__tail₁(x1) = a__tail₁(x1)
0 = 0

Lexicographic Path Order [19].
Precedence:

[MARK₁, A_{_{TAIL_1]}} > [tail₁, mark₁, a_{_{tail_1]}} > a_{_zeros} > cons₂
[MARK₁, A_{_{TAIL_1]}} > [tail₁, mark₁, a_{_{tail_1]}} > a_{_zeros} > zeros
[MARK₁, A_{_{TAIL_1]}} > [tail₁, mark₁, a_{_{tail_1]}} > a_{_zeros} > 0

The following usable rules [14] were oriented:

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

↳ 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))

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.