Termination w.r.t. Q proof of TRS/TRCSREx2_Luc03b

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:

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ PisEmptyProof

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

fst₂(0, Z) -> nil
fst₂(s, cons₁(Y)) -> cons₁(Y)
from₁(X) -> cons₁(X)
add₂(0, X) -> X
add₂(s, Y) -> s
len₁(nil) -> 0
len₁(cons₁(X)) -> s

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.