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:

active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
active(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(half(X)) → half(active(X))
terms(mark(X)) → mark(terms(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
recip(mark(X)) → mark(recip(X))
sqr(mark(X)) → mark(sqr(X))
s(mark(X)) → mark(s(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
dbl(mark(X)) → mark(dbl(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
half(mark(X)) → mark(half(X))
proper(terms(X)) → terms(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(recip(X)) → recip(proper(X))
proper(sqr(X)) → sqr(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(dbl(X)) → dbl(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(half(X)) → half(proper(X))
terms(ok(X)) → ok(terms(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
recip(ok(X)) → ok(recip(X))
sqr(ok(X)) → ok(sqr(X))
s(ok(X)) → ok(s(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
dbl(ok(X)) → ok(dbl(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
half(ok(X)) → ok(half(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) QTRSToCSRProof (EQUIVALENT transformation)

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

active(terms(N)) → mark(cons(recip(sqr(N)), terms(s(N))))
active(sqr(0)) → mark(0)
active(sqr(s(X))) → mark(s(add(sqr(X), dbl(X))))
active(dbl(0)) → mark(0)
active(dbl(s(X))) → mark(s(s(dbl(X))))
active(add(0, X)) → mark(X)
active(add(s(X), Y)) → mark(s(add(X, Y)))
active(first(0, X)) → mark(nil)
active(first(s(X), cons(Y, Z))) → mark(cons(Y, first(X, Z)))
active(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
active(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(half(X)) → half(active(X))
terms(mark(X)) → mark(terms(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
recip(mark(X)) → mark(recip(X))
sqr(mark(X)) → mark(sqr(X))
s(mark(X)) → mark(s(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
dbl(mark(X)) → mark(dbl(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
half(mark(X)) → mark(half(X))
proper(terms(X)) → terms(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(recip(X)) → recip(proper(X))
proper(sqr(X)) → sqr(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(dbl(X)) → dbl(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(half(X)) → half(proper(X))
terms(ok(X)) → ok(terms(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
recip(ok(X)) → ok(recip(X))
sqr(ok(X)) → ok(sqr(X))
s(ok(X)) → ok(s(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
dbl(ok(X)) → ok(dbl(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
half(ok(X)) → ok(half(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
Special symbols used for the transformation (see [GM04]):
top: top, active: active, mark: mark, ok: ok, proper: proper
The replacement map contains the following entries:

terms: {1}
cons: {1}
recip: {1}
sqr: {1}
s: {1}
0: empty set
add: {1, 2}
dbl: {1}
first: {1, 2}
nil: empty set
half: {1}
The QTRS contained all rules created by the complete Giesl-Middeldorp transformation. Therefore, the inverse transformation is complete (and sound).

(2) Obligation:

Context-sensitive rewrite system:
The TRS R consists of the following rules:

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

The replacement map contains the following entries:

terms: {1}
cons: {1}
recip: {1}
sqr: {1}
s: {1}
0: empty set
add: {1, 2}
dbl: {1}
first: {1, 2}
nil: empty set
half: {1}

(3) CSDependencyPairsProof (EQUIVALENT transformation)

Using Improved CS-DPs [LPAR08] we result in the following initial Q-CSDP problem.

(4) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half, SQR, TERMS, ADD, DBL, HALF} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The ordinary context-sensitive dependency pairs DP_o are:

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)), terms(s(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, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

Q is empty.

(5) QCSDependencyGraphProof (EQUIVALENT transformation)

The approximation of the Context-Sensitive Dependency Graph [LPAR08] contains 4 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half, HALF} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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)), terms(s(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, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

Q is empty.

(8) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].

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.
none
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  x1

Subterm Order

(9) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

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

Q is empty.

(10) PIsEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half, DBL} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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)), terms(s(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, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

Q is empty.

(13) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].

The following pairs can be oriented strictly and are deleted.

DBL(s(X)) → DBL(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
DBL(x1)  =  x1

Subterm Order

(14) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

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

Q is empty.

(15) PIsEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half, ADD} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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)), terms(s(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, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

Q is empty.

(18) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].

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.
none
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2)  =  x1

Subterm Order

(19) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

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

Q is empty.

(20) PIsEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half, SQR} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

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)), terms(s(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, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

Q is empty.

(23) QCSDPSubtermProof (EQUIVALENT transformation)

We use the subterm processor [DA_EMMES].

The following pairs can be oriented strictly and are deleted.

SQR(s(X)) → SQR(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SQR(x1)  =  x1

Subterm Order

(24) Obligation:

Q-restricted context-sensitive dependency pair problem:
The symbols in {terms, recip, sqr, s, add, dbl, first, half} are replacing on all positions.
For all symbols f in {cons} we have µ(f) = {1}.

The TRS P consists of the following rules:
none

The TRS R consists of the following rules:

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

Q is empty.

(25) PIsEmptyProof (EQUIVALENT transformation)

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

(26) TRUE