Termination w.r.t. Q proof of /tmp/tmpjEvgaG/Ex26_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:

terms(N) → cons(recip(sqr(N)))
sqr(0) → 0
sqr(s) → s
dbl(0) → 0
dbl(s) → s
add(0, X) → X
add(s, Y) → s
first(0, X) → nil
first(s, cons(Y)) → cons(Y)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

terms(N) → cons(recip(sqr(N)))
sqr(0) → 0
sqr(s) → s
dbl(0) → 0
dbl(s) → s
add(0, X) → X
add(s, Y) → s
first(0, X) → nil
first(s, cons(Y)) → cons(Y)

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:

TERMS(N) → SQR(N)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)))
sqr(0) → 0
sqr(s) → s
dbl(0) → 0
dbl(s) → s
add(0, X) → X
add(s, Y) → s
first(0, X) → nil
first(s, cons(Y)) → cons(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:

TERMS(N) → SQR(N)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)))
sqr(0) → 0
sqr(s) → s
dbl(0) → 0
dbl(s) → s
add(0, X) → X
add(s, Y) → s
first(0, X) → nil
first(s, cons(Y)) → cons(Y)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.