Termination w.r.t. Q proof of TRS/TRCSREx26_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:

terms₁(N) -> cons₁(recip₁(sqr₁(N)))
sqr₁(0) -> 0
sqr₁(s) -> s
dbl₁(0) -> 0
dbl₁(s) -> s
add₂(0, X) -> X
add₂(s, Y) -> s
first₂(0, X) -> nil
first₂(s, cons₁(Y)) -> cons₁(Y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

terms₁(N) -> cons₁(recip₁(sqr₁(N)))
sqr₁(0) -> 0
sqr₁(s) -> s
dbl₁(0) -> 0
dbl₁(s) -> s
add₂(0, X) -> X
add₂(s, Y) -> s
first₂(0, X) -> nil
first₂(s, cons₁(Y)) -> cons₁(Y)

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:

TERMS₁(N) -> SQR₁(N)

The TRS R consists of the following rules:

terms₁(N) -> cons₁(recip₁(sqr₁(N)))
sqr₁(0) -> 0
sqr₁(s) -> s
dbl₁(0) -> 0
dbl₁(s) -> s
add₂(0, X) -> X
add₂(s, Y) -> s
first₂(0, X) -> nil
first₂(s, cons₁(Y)) -> cons₁(Y)

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:

TERMS₁(N) -> SQR₁(N)

The TRS R consists of the following rules:

terms₁(N) -> cons₁(recip₁(sqr₁(N)))
sqr₁(0) -> 0
sqr₁(s) -> s
dbl₁(0) -> 0
dbl₁(s) -> s
add₂(0, X) -> X
add₂(s, Y) -> s
first₂(0, X) -> nil
first₂(s, cons₁(Y)) -> cons₁(Y)

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.