Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

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

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TERMS(N) → SQR(N)
SQR(s(X)) → ADD(sqr(X), dbl(X))
SQR(s(X)) → SQR(X)
SQR(s(X)) → DBL(X)
DBL(s(X)) → DBL(X)
ADD(s(X), Y) → ADD(X, Y)
HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

HALF(s(s(X))) → HALF(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  HALF(x1)
s(x1)  =  s(x1)
terms(x1)  =  terms
cons(x1)  =  cons
recip(x1)  =  x1
sqr(x1)  =  sqr(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first(x1)
nil  =  nil
half(x1)  =  half(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

terms > [cons, first₁] > nil
terms > [sqr₁, dbl₁] > add₂ > s₁ > 0
half₁ > s₁ > 0

Status:

add₂: [1,2]
HALF₁: [1]
cons: []
half₁: [1]
sqr₁: [1]
dbl₁: [1]
s₁: [1]
0: []
first₁: [1]
nil: []
terms: []

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

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

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(s(X), Y) → ADD(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2)  =  ADD(x1)
s(x1)  =  s(x1)
terms(x1)  =  terms
cons(x1)  =  cons
recip(x1)  =  recip
sqr(x1)  =  sqr(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first
nil  =  nil
half(x1)  =  half(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

ADD₁ > [0, nil]
terms > cons > [0, nil]
recip > [0, nil]
sqr₁ > add₂ > s₁ > [0, nil]
sqr₁ > dbl₁ > s₁ > [0, nil]
first > cons > [0, nil]
half₁ > s₁ > [0, nil]

Status:

sqr₁: [1]
ADD₁: [1]
recip: []
0: []
first: []
add₂: [2,1]
cons: []
half₁: [1]
dbl₁: [1]
s₁: [1]
nil: []
terms: []

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(s(X)) → DBL(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DBL(x1)  =  DBL(x1)
s(x1)  =  s(x1)
terms(x1)  =  terms(x1)
cons(x1)  =  cons
recip(x1)  =  x1
sqr(x1)  =  sqr(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first
nil  =  nil
half(x1)  =  half(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

DBL₁ > nil
terms₁ > cons > nil
sqr₁ > 0 > nil
sqr₁ > add₂ > s₁ > cons > nil
sqr₁ > add₂ > s₁ > half₁ > nil
sqr₁ > dbl₁ > s₁ > cons > nil
sqr₁ > dbl₁ > s₁ > half₁ > nil
first > cons > nil

Status:

DBL₁: [1]
add₂: [2,1]
cons: []
half₁: [1]
sqr₁: [1]
dbl₁: [1]
s₁: [1]
0: []
terms₁: [1]
nil: []
first: []

The following usable rules [FROCOS05] were oriented:

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

(17) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) TRUE

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

SQR(s(X)) → SQR(X)

The TRS R consists of the following rules:

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

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

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SQR(s(X)) → SQR(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1)  =  SQR(x1)
s(x1)  =  s(x1)
terms(x1)  =  terms(x1)
cons(x1)  =  cons
recip(x1)  =  x1
sqr(x1)  =  sqr(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first
nil  =  nil
half(x1)  =  half(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

SQR₁ > nil
terms₁ > cons > nil
sqr₁ > 0 > nil
sqr₁ > add₂ > s₁ > cons > nil
sqr₁ > add₂ > s₁ > half₁ > nil
sqr₁ > dbl₁ > s₁ > cons > nil
sqr₁ > dbl₁ > s₁ > half₁ > nil
first > cons > nil

Status:

add₂: [2,1]
cons: []
half₁: [1]
sqr₁: [1]
SQR₁: [1]
dbl₁: [1]
s₁: [1]
0: []
terms₁: [1]
nil: []
first: []

The following usable rules [FROCOS05] were oriented:

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

(22) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) TRUE