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:

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

  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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)

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s)
dbl(0)
dbl(s)
add(0, x0)
add(s, x0)
first(0, x0)
first(s, cons(x0))

Using Dependency Pairs [1,13] 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)

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s)
dbl(0)
dbl(s)
add(0, x0)
add(s, x0)
first(0, x0)
first(s, cons(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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)

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s)
dbl(0)
dbl(s)
add(0, x0)
add(s, x0)
first(0, x0)
first(s, cons(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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)

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s)
dbl(0)
dbl(s)
add(0, x0)
add(s, x0)
first(0, x0)
first(s, cons(x0))

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.