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:

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

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:

F(empty, cons(a, k)) → F(cons(a, k), k)
F(cons(a, k), y) → F(y, k)

The TRS R consists of the following rules:

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

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:

F(empty, cons(a, k)) → F(cons(a, k), k)
F(cons(a, k), y) → F(y, k)

The TRS R consists of the following rules:

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

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.


F(empty, cons(a, k)) → F(cons(a, k), k)
F(cons(a, k), y) → F(y, k)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(cons(x1, x2)) = 3 + (2)x_1 + (4)x_2   
POL(empty) = 0   
POL(F(x1, x2)) = (3)x_1 + (4)x_2   
The value of delta used in the strict ordering is 3.
The following usable rules [17] were oriented: none



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

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

f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

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.