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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s)
dbl1(0)
dbl1(s)
add2(0, x0)
add2(s, x0)
first2(0, x0)
first2(s, cons1(x0))


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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s)
dbl1(0)
dbl1(s)
add2(0, x0)
add2(s, x0)
first2(0, x0)
first2(s, cons1(x0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)

The set Q consists of the following terms:

terms1(x0)
sqr1(0)
sqr1(s)
dbl1(0)
dbl1(s)
add2(0, x0)
add2(s, x0)
first2(0, x0)
first2(s, cons1(x0))

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