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__2nd(cons(X, cons(Y, Z))) → mark(Y)
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__2nd(cons(X, cons(Y, Z))) → mark(Y)
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

A__2ND(cons(X, cons(Y, Z))) → MARK(Y)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(2nd(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(2nd(X)) → A__2ND(mark(X))
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__2nd(cons(X, cons(Y, Z))) → mark(Y)
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__2nd(X) → 2nd(X)
a__from(X) → from(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:

A__2ND(cons(X, cons(Y, Z))) → MARK(Y)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(2nd(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(2nd(X)) → A__2ND(mark(X))
A__FROM(X) → MARK(X)

The TRS R consists of the following rules:

a__2nd(cons(X, cons(Y, Z))) → mark(Y)
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


A__2ND(cons(X, cons(Y, Z))) → MARK(Y)
MARK(from(X)) → MARK(X)
MARK(from(X)) → A__FROM(mark(X))
MARK(s(X)) → MARK(X)
MARK(2nd(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(2nd(X)) → A__2ND(mark(X))
A__FROM(X) → MARK(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(A__2ND(x1)) = x_1   
POL(cons(x1, x2)) = 1/2 + (2)x_1 + (1/4)x_2   
POL(from(x1)) = 9/4 + (3)x_1   
POL(MARK(x1)) = 1/4 + (1/2)x_1   
POL(A__FROM(x1)) = 1/2 + (3/2)x_1   
POL(a__2nd(x1)) = 1/4 + (3)x_1   
POL(2nd(x1)) = 1/4 + (3)x_1   
POL(mark(x1)) = x_1   
POL(s(x1)) = 1/2 + x_1   
POL(a__from(x1)) = 9/4 + (3)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

a__from(X) → cons(mark(X), from(s(X)))
a__2nd(cons(X, cons(Y, Z))) → mark(Y)
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__2nd(X) → 2nd(X)
a__from(X) → from(X)



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__2nd(cons(X, cons(Y, Z))) → mark(Y)
a__from(X) → cons(mark(X), from(s(X)))
mark(2nd(X)) → a__2nd(mark(X))
mark(from(X)) → a__from(mark(X))
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(s(X)) → s(mark(X))
a__2nd(X) → 2nd(X)
a__from(X) → from(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.