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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(terms(N)) → CONS(recip(sqr(N)), terms(s(N)))
ACTIVE(terms(N)) → RECIP(sqr(N))
ACTIVE(terms(N)) → SQR(N)
ACTIVE(terms(N)) → TERMS(s(N))
ACTIVE(terms(N)) → S(N)
ACTIVE(sqr(s(X))) → S(add(sqr(X), dbl(X)))
ACTIVE(sqr(s(X))) → ADD(sqr(X), dbl(X))
ACTIVE(sqr(s(X))) → SQR(X)
ACTIVE(sqr(s(X))) → DBL(X)
ACTIVE(dbl(s(X))) → S(s(dbl(X)))
ACTIVE(dbl(s(X))) → S(dbl(X))
ACTIVE(dbl(s(X))) → DBL(X)
ACTIVE(add(s(X), Y)) → S(add(X, Y))
ACTIVE(add(s(X), Y)) → ADD(X, Y)
ACTIVE(first(s(X), cons(Y, Z))) → CONS(Y, first(X, Z))
ACTIVE(first(s(X), cons(Y, Z))) → FIRST(X, Z)
ACTIVE(terms(X)) → TERMS(active(X))
ACTIVE(terms(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(recip(X)) → RECIP(active(X))
ACTIVE(recip(X)) → ACTIVE(X)
ACTIVE(sqr(X)) → SQR(active(X))
ACTIVE(sqr(X)) → ACTIVE(X)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(add(X1, X2)) → ADD(active(X1), X2)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ADD(X1, active(X2))
ACTIVE(add(X1, X2)) → ACTIVE(X2)
ACTIVE(dbl(X)) → DBL(active(X))
ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(first(X1, X2)) → FIRST(active(X1), X2)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → FIRST(X1, active(X2))
ACTIVE(first(X1, X2)) → ACTIVE(X2)
TERMS(mark(X)) → TERMS(X)
CONS(mark(X1), X2) → CONS(X1, X2)
RECIP(mark(X)) → RECIP(X)
SQR(mark(X)) → SQR(X)
S(mark(X)) → S(X)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(X1, mark(X2)) → ADD(X1, X2)
DBL(mark(X)) → DBL(X)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(X1, mark(X2)) → FIRST(X1, X2)
PROPER(terms(X)) → TERMS(proper(X))
PROPER(terms(X)) → PROPER(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(recip(X)) → RECIP(proper(X))
PROPER(recip(X)) → PROPER(X)
PROPER(sqr(X)) → SQR(proper(X))
PROPER(sqr(X)) → PROPER(X)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(add(X1, X2)) → ADD(proper(X1), proper(X2))
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(dbl(X)) → DBL(proper(X))
PROPER(dbl(X)) → PROPER(X)
PROPER(first(X1, X2)) → FIRST(proper(X1), proper(X2))
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)
TERMS(ok(X)) → TERMS(X)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
RECIP(ok(X)) → RECIP(X)
SQR(ok(X)) → SQR(X)
S(ok(X)) → S(X)
ADD(ok(X1), ok(X2)) → ADD(X1, X2)
DBL(ok(X)) → DBL(X)
FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 11 SCCs with 36 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(mark(X1), X2) → FIRST(X1, X2)
FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

FIRST(X1, mark(X2)) → FIRST(X1, X2)
FIRST(ok(X1), ok(X2)) → FIRST(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

[active₁, first₂, proper₁] > add₂ > [ok₁, terms₁, cons₁, recip₁] > [mark₁, top]
[0, nil] > [ok₁, terms₁, cons₁, recip₁] > [mark₁, top]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

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

FIRST(mark(X1), X2) → FIRST(X1, X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

FIRST(mark(X1), X2) → FIRST(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

[active₁, add₂] > recip₁ > [mark₁, ok, top]
[active₁, add₂] > [s₁, first₂] > [mark₁, ok, top]
[0, nil] > [mark₁, ok, top]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(9) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

DBL(ok(X)) → DBL(X)
DBL(mark(X)) → DBL(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(mark(X)) → DBL(X)

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

Lexicographic Path Order [LPO].
Precedence:

top > [active₁, nil] > 0
top > [active₁, nil] > add₂ > [mark₁, s₁]
top > [active₁, nil] > first₂ > [mark₁, s₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(14) Obligation:

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

DBL(ok(X)) → DBL(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

DBL(ok(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)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
terms(x1)  =  terms(x1)
mark(x1)  =  mark
cons(x1, x2)  =  x1
recip(x1)  =  x1
sqr(x1)  =  x1
s(x1)  =  x1
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  x1
first(x1, x2)  =  first(x2)
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:

active₁ > [terms₁, add₂, first₁, proper₁] > [DBL₁, ok₁] > top
active₁ > [terms₁, add₂, first₁, proper₁] > mark > top
active₁ > [terms₁, add₂, first₁, proper₁] > 0 > top
active₁ > nil > [DBL₁, ok₁] > top

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(16) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(mark(X1), X2) → ADD(X1, X2)
ADD(ok(X1), ok(X2)) → ADD(X1, X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(X1, mark(X2)) → ADD(X1, X2)
ADD(ok(X1), ok(X2)) → ADD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

active₁ > [0, first₂, proper₁] > add₂ > [ok₁, terms₁, cons₁, recip₁] > [mark₁, top]
active₁ > nil > [ok₁, terms₁, cons₁, recip₁] > [mark₁, top]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(21) Obligation:

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

ADD(mark(X1), X2) → ADD(X1, X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ADD(mark(X1), X2) → ADD(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

[active₁, add₂] > recip₁ > [mark₁, ok, top]
[active₁, add₂] > [s₁, first₂] > [mark₁, ok, top]
[0, nil] > [mark₁, ok, top]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(23) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

S(ok(X)) → S(X)
S(mark(X)) → S(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

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

Lexicographic Path Order [LPO].
Precedence:

top > [active₁, nil] > 0
top > [active₁, nil] > add₂ > [mark₁, s₁]
top > [active₁, nil] > first₂ > [mark₁, s₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(28) Obligation:

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

S(ok(X)) → S(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(ok(X)) → S(X)

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

Lexicographic Path Order [LPO].
Precedence:

active₁ > [terms₁, add₂, first₁, proper₁] > [S₁, ok₁] > top
active₁ > [terms₁, add₂, first₁, proper₁] > mark > top
active₁ > [terms₁, add₂, first₁, proper₁] > 0 > top
active₁ > nil > [S₁, ok₁] > top

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(30) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

SQR(ok(X)) → SQR(X)
SQR(mark(X)) → SQR(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SQR(mark(X)) → SQR(X)

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

Lexicographic Path Order [LPO].
Precedence:

top > [active₁, nil] > 0
top > [active₁, nil] > add₂ > [mark₁, s₁]
top > [active₁, nil] > first₂ > [mark₁, s₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(35) Obligation:

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

SQR(ok(X)) → SQR(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SQR(ok(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)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
terms(x1)  =  terms(x1)
mark(x1)  =  mark
cons(x1, x2)  =  x1
recip(x1)  =  x1
sqr(x1)  =  x1
s(x1)  =  x1
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  x1
first(x1, x2)  =  first(x2)
nil  =  nil
proper(x1)  =  proper(x1)
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:

active₁ > [terms₁, add₂, first₁, proper₁] > [SQR₁, ok₁] > top
active₁ > [terms₁, add₂, first₁, proper₁] > mark > top
active₁ > [terms₁, add₂, first₁, proper₁] > 0 > top
active₁ > nil > [SQR₁, ok₁] > top

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(37) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

RECIP(ok(X)) → RECIP(X)
RECIP(mark(X)) → RECIP(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

RECIP(mark(X)) → RECIP(X)

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

Lexicographic Path Order [LPO].
Precedence:

top > [active₁, nil] > 0
top > [active₁, nil] > add₂ > [mark₁, s₁]
top > [active₁, nil] > first₂ > [mark₁, s₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(42) Obligation:

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

RECIP(ok(X)) → RECIP(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

RECIP(ok(X)) → RECIP(X)

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

Lexicographic Path Order [LPO].
Precedence:

active₁ > [terms₁, add₂, first₁, proper₁] > [RECIP₁, ok₁] > top
active₁ > [terms₁, add₂, first₁, proper₁] > mark > top
active₁ > [terms₁, add₂, first₁, proper₁] > 0 > top
active₁ > nil > [RECIP₁, ok₁] > top

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(44) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(48) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(mark(X1), X2) → CONS(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

CONS₂ > [mark₁, cons₁]
top > [active₁, first₂] > recip₁ > [mark₁, cons₁]
top > [active₁, first₂] > [0, nil] > [mark₁, cons₁]
top > [active₁, first₂] > add₂ > [mark₁, cons₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(49) Obligation:

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

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

CONS(ok(X1), ok(X2)) → CONS(X1, X2)

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

Lexicographic Path Order [LPO].
Precedence:

[CONS₁, ok₁, active₁, cons₁, s₁, first₁, proper₁] > 0 > [nil, top]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(51) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(52) PisEmptyProof (EQUIVALENT transformation)

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

(53) TRUE

(54) Obligation:

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

TERMS(ok(X)) → TERMS(X)
TERMS(mark(X)) → TERMS(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TERMS(mark(X)) → TERMS(X)

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

Lexicographic Path Order [LPO].
Precedence:

top > [active₁, nil] > 0
top > [active₁, nil] > add₂ > [mark₁, s₁]
top > [active₁, nil] > first₂ > [mark₁, s₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(56) Obligation:

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

TERMS(ok(X)) → TERMS(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(57) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TERMS(ok(X)) → TERMS(X)

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

Lexicographic Path Order [LPO].
Precedence:

active₁ > [terms₁, add₂, first₁, proper₁] > [TERMS₁, ok₁] > top
active₁ > [terms₁, add₂, first₁, proper₁] > mark > top
active₁ > [terms₁, add₂, first₁, proper₁] > 0 > top
active₁ > nil > [TERMS₁, ok₁] > top

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(58) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(59) PisEmptyProof (EQUIVALENT transformation)

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

(60) TRUE

(61) Obligation:

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

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(terms(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(recip(X)) → PROPER(X)
PROPER(sqr(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(dbl(X)) → PROPER(X)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(terms(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(sqr(X)) → PROPER(X)
PROPER(add(X1, X2)) → PROPER(X1)
PROPER(add(X1, X2)) → PROPER(X2)
PROPER(first(X1, X2)) → PROPER(X1)
PROPER(first(X1, X2)) → PROPER(X2)

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

Lexicographic Path Order [LPO].
Precedence:

active₁ > terms₁ > sqr₁ > [0, nil, ok, top] > [cons₂, add₂, mark]
active₁ > first₂ > [0, nil, ok, top] > [cons₂, add₂, mark]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(63) Obligation:

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

PROPER(recip(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(dbl(X)) → PROPER(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(recip(X)) → PROPER(X)

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

Lexicographic Path Order [LPO].
Precedence:

top > [0, proper₁] > terms > [PROPER₁, recip₁, active₁] > mark
top > [0, proper₁] > cons₁ > [PROPER₁, recip₁, active₁] > mark
top > [0, proper₁] > sqr > [PROPER₁, recip₁, active₁] > mark
top > [0, proper₁] > nil > mark

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(65) Obligation:

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

PROPER(s(X)) → PROPER(X)
PROPER(dbl(X)) → PROPER(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(66) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(dbl(X)) → PROPER(X)

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

Lexicographic Path Order [LPO].
Precedence:

active₁ > [dbl₁, proper₁] > terms₁ > [cons, recip, nil, ok, top]
active₁ > [dbl₁, proper₁] > sqr₁ > [cons, recip, nil, ok, top]
active₁ > [dbl₁, proper₁] > 0 > [cons, recip, nil, ok, top]
active₁ > [dbl₁, proper₁] > first₁ > [cons, recip, nil, ok, top]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(67) Obligation:

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

PROPER(s(X)) → PROPER(X)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)

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

Lexicographic Path Order [LPO].
Precedence:

0 > [s₁, active₁, nil] > [PROPER₁, terms, mark]
dbl > [s₁, active₁, nil] > [PROPER₁, terms, mark]
first > [s₁, active₁, nil] > [PROPER₁, terms, mark]
top > [PROPER₁, terms, mark]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(69) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(70) PisEmptyProof (EQUIVALENT transformation)

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

(71) TRUE

(72) Obligation:

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

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(terms(X)) → ACTIVE(X)
ACTIVE(recip(X)) → ACTIVE(X)
ACTIVE(sqr(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X2)
ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X2)

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(73) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(terms(X)) → ACTIVE(X)
ACTIVE(recip(X)) → ACTIVE(X)
ACTIVE(sqr(X)) → ACTIVE(X)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(add(X1, X2)) → ACTIVE(X1)
ACTIVE(add(X1, X2)) → ACTIVE(X2)
ACTIVE(dbl(X)) → ACTIVE(X)
ACTIVE(first(X1, X2)) → ACTIVE(X1)
ACTIVE(first(X1, X2)) → ACTIVE(X2)

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

Lexicographic Path Order [LPO].
Precedence:

[active₁, 0] > sqr₁ > [ACTIVE₁, add₂] > [mark, nil, ok, top] > cons₁ > first₂ > s₁
[active₁, 0] > sqr₁ > [ACTIVE₁, add₂] > [mark, nil, ok, top] > terms₁ > s₁
[active₁, 0] > sqr₁ > [ACTIVE₁, add₂] > [mark, nil, ok, top] > recip₁ > s₁
[active₁, 0] > sqr₁ > [ACTIVE₁, add₂] > [mark, nil, ok, top] > dbl₁ > s₁

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(74) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(75) PisEmptyProof (EQUIVALENT transformation)

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

(76) TRUE

(77) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(78) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(mark(X)) → TOP(proper(X))

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

Lexicographic Path Order [LPO].
Precedence:

TOP₁ > [mark₁, s₁]
terms₁ > [cons₁, first₂] > nil > [mark₁, s₁]
terms₁ > [sqr₁, dbl₁] > add₂ > [mark₁, s₁]
0 > nil > [mark₁, s₁]
top > [mark₁, s₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(79) Obligation:

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

TOP(ok(X)) → TOP(active(X))

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(80) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(ok(X)) → TOP(active(X))

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

Lexicographic Path Order [LPO].
Precedence:

0 > [TOP₁, ok₁, mark, recip₁, sqr₁, proper₁]
nil > [TOP₁, ok₁, mark, recip₁, sqr₁, proper₁]
top > [TOP₁, ok₁, mark, recip₁, sqr₁, proper₁]

The following usable rules [FROCOS05] were oriented:

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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(81) Obligation:

Q DP problem:
P is empty.
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(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))
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))
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)
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))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(82) PisEmptyProof (EQUIVALENT transformation)

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

(83) TRUE