Termination w.r.t. Q proof of /tmp/tmpsm3iou/Ex2_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:

fst(0, Z) → nil
fst(s, cons(Y)) → cons(Y)
from(X) → cons(X)
add(0, X) → X
add(s, Y) → s
len(nil) → 0
len(cons(X)) → s

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fst(0, Z) → nil
fst(s, cons(Y)) → cons(Y)
from(X) → cons(X)
add(0, X) → X
add(s, Y) → s
len(nil) → 0
len(cons(X)) → s

Q is empty.

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

fst(0, Z) → nil
fst(s, cons(Y)) → cons(Y)
from(X) → cons(X)
add(0, X) → X
add(s, Y) → s
len(nil) → 0
len(cons(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:

fst(0, Z) → nil
fst(s, cons(Y)) → cons(Y)
from(X) → cons(X)
add(0, X) → X
add(s, Y) → s
len(nil) → 0
len(cons(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.