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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 PisEmptyProof (⇔)
    ↳18 TRUE
  ↳19 QDP
    ↳20 QDPOrderProof (⇔)
    ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 PisEmptyProof (⇔)
    ↳25 TRUE
  ↳26 QDP
    ↳27 QDPOrderProof (⇔)
    ↳28 QDP
    ↳29 QDPOrderProof (⇔)
    ↳30 QDP
    ↳31 PisEmptyProof (⇔)
    ↳32 TRUE
  ↳33 QDP
    ↳34 QDPOrderProof (⇔)
    ↳35 QDP
    ↳36 QDPOrderProof (⇔)
    ↳37 QDP
    ↳38 PisEmptyProof (⇔)
    ↳39 TRUE
  ↳40 QDP
    ↳41 QDPOrderProof (⇔)
    ↳42 QDP
    ↳43 QDPOrderProof (⇔)
    ↳44 QDP
    ↳45 PisEmptyProof (⇔)
    ↳46 TRUE
  ↳47 QDP
    ↳48 QDPOrderProof (⇔)
    ↳49 QDP
    ↳50 QDPOrderProof (⇔)
    ↳51 QDP
    ↳52 PisEmptyProof (⇔)
    ↳53 TRUE
  ↳54 QDP
    ↳55 QDPOrderProof (⇔)
    ↳56 QDP
    ↳57 QDPOrderProof (⇔)
    ↳58 QDP
    ↳59 PisEmptyProof (⇔)
    ↳60 TRUE
  ↳61 QDP
    ↳62 QDPOrderProof (⇔)
    ↳63 QDP
    ↳64 QDPOrderProof (⇔)
    ↳65 QDP
    ↳66 PisEmptyProof (⇔)
    ↳67 TRUE
  ↳68 QDP
    ↳69 QDPOrderProof (⇔)
    ↳70 QDP
    ↳71 QDPOrderProof (⇔)
    ↳72 QDP
    ↳73 PisEmptyProof (⇔)
    ↳74 TRUE
  ↳75 QDP
    ↳76 QDPOrderProof (⇔)
    ↳77 QDP
    ↳78 QDPOrderProof (⇔)
    ↳79 QDP
    ↳80 PisEmptyProof (⇔)
    ↳81 TRUE
  ↳82 QDP
    ↳83 QDPOrderProof (⇔)
    ↳84 QDP
    ↳85 QDPOrderProof (⇔)
    ↳86 QDP
    ↳87 PisEmptyProof (⇔)
    ↳88 TRUE

(0) Obligation:

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

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

Q is empty.

(1) 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(half(s(s(X)))) → S(half(X))
ACTIVE(half(s(s(X)))) → HALF(X)
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)
ACTIVE(half(X)) → HALF(active(X))
ACTIVE(half(X)) → ACTIVE(X)
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)
HALF(mark(X)) → HALF(X)
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)
PROPER(half(X)) → HALF(proper(X))
PROPER(half(X)) → PROPER(X)
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)
HALF(ok(X)) → HALF(X)
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(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
active(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(half(X)) → half(active(X))
terms(mark(X)) → mark(terms(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
recip(mark(X)) → mark(recip(X))
sqr(mark(X)) → mark(sqr(X))
s(mark(X)) → mark(s(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
dbl(mark(X)) → mark(dbl(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
half(mark(X)) → mark(half(X))
proper(terms(X)) → terms(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(recip(X)) → recip(proper(X))
proper(sqr(X)) → sqr(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(dbl(X)) → dbl(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(half(X)) → half(proper(X))
terms(ok(X)) → ok(terms(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
recip(ok(X)) → ok(recip(X))
sqr(ok(X)) → ok(sqr(X))
s(ok(X)) → ok(s(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
dbl(ok(X)) → ok(dbl(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
half(ok(X)) → ok(half(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 12 SCCs with 40 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HALF(ok(X)) → HALF(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

Q is empty.
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.


HALF(mark(X)) → HALF(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HALF(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


FIRST(mark(X1), 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)  =  x1
mark(x1)  =  mark(x1)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

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

Q is empty.
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.


FIRST(X1, mark(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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


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)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

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

Q is empty.
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.


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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


ADD(mark(X1), 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)  =  x1
mark(x1)  =  mark(x1)
ok(x1)  =  ok(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

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

Q is empty.
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.


ADD(X1, mark(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)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


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)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

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

Q is empty.
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.


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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


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)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

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

Q is empty.
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.


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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


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)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(49) Obligation:

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

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

Q is empty.
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.


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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


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)  =  x1
ok(x1)  =  x1
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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.


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)

Lexicographic Path Order [LPO].
Precedence:
ok1 > CONS1


The following usable rules [FROCOS05] were oriented: none

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

Q is empty.
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:

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

Q is empty.
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.


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)  =  x1
ok(x1)  =  ok(x1)
mark(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(63) Obligation:

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

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

Q is empty.
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.


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
mark(x1)  =  mark(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

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

(66) PisEmptyProof (EQUIVALENT transformation)

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

(67) TRUE

(68) 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)
PROPER(half(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(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
active(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(half(X)) → half(active(X))
terms(mark(X)) → mark(terms(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
recip(mark(X)) → mark(recip(X))
sqr(mark(X)) → mark(sqr(X))
s(mark(X)) → mark(s(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
dbl(mark(X)) → mark(dbl(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
half(mark(X)) → mark(half(X))
proper(terms(X)) → terms(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(recip(X)) → recip(proper(X))
proper(sqr(X)) → sqr(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(dbl(X)) → dbl(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(half(X)) → half(proper(X))
terms(ok(X)) → ok(terms(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
recip(ok(X)) → ok(recip(X))
sqr(ok(X)) → ok(sqr(X))
s(ok(X)) → ok(s(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
dbl(ok(X)) → ok(dbl(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
half(ok(X)) → ok(half(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(69) 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(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 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)  =  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)
half(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(70) Obligation:

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

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

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

(71) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(half(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
half(x1)  =  half(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

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

(73) PisEmptyProof (EQUIVALENT transformation)

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

(74) TRUE

(75) 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)
ACTIVE(half(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(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
active(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(half(X)) → half(active(X))
terms(mark(X)) → mark(terms(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
recip(mark(X)) → mark(recip(X))
sqr(mark(X)) → mark(sqr(X))
s(mark(X)) → mark(s(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
dbl(mark(X)) → mark(dbl(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
half(mark(X)) → mark(half(X))
proper(terms(X)) → terms(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(recip(X)) → recip(proper(X))
proper(sqr(X)) → sqr(proper(X))
proper(s(X)) → s(proper(X))
proper(0) → ok(0)
proper(add(X1, X2)) → add(proper(X1), proper(X2))
proper(dbl(X)) → dbl(proper(X))
proper(first(X1, X2)) → first(proper(X1), proper(X2))
proper(nil) → ok(nil)
proper(half(X)) → half(proper(X))
terms(ok(X)) → ok(terms(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
recip(ok(X)) → ok(recip(X))
sqr(ok(X)) → ok(sqr(X))
s(ok(X)) → ok(s(X))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
dbl(ok(X)) → ok(dbl(X))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
half(ok(X)) → ok(half(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(76) 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)  =  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)
half(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(77) Obligation:

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

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

Q is empty.
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.


ACTIVE(half(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  x1
half(x1)  =  half(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

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

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

(80) PisEmptyProof (EQUIVALENT transformation)

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

(81) TRUE

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

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

(83) 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
half(x1)  =  half(x1)

Lexicographic Path Order [LPO].
Precedence:
terms1 > [sqr1, dbl1] > add2 > [s1, first2] > half1 > [TOP1, mark1, cons1, 0, nil]


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

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

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

(85) 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)  =  x1
sqr(x1)  =  x1
s(x1)  =  x1
0  =  0
add(x1, x2)  =  x1
dbl(x1)  =  x1
first(x1, x2)  =  x1
nil  =  nil
half(x1)  =  x1

Lexicographic Path Order [LPO].
Precedence:
TOP1 > [mark, 0, nil]
ok1 > [mark, 0, nil]


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(half(0)) → mark(0)
active(half(s(0))) → mark(0)
active(half(s(s(X)))) → mark(s(half(X)))
active(half(dbl(X))) → mark(X)
active(terms(X)) → terms(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(recip(X)) → recip(active(X))
active(sqr(X)) → sqr(active(X))
active(s(X)) → s(active(X))
active(add(X1, X2)) → add(active(X1), X2)
active(add(X1, X2)) → add(X1, active(X2))
active(dbl(X)) → dbl(active(X))
active(first(X1, X2)) → first(active(X1), X2)
active(first(X1, X2)) → first(X1, active(X2))
active(half(X)) → half(active(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
terms(mark(X)) → mark(terms(X))
terms(ok(X)) → ok(terms(X))
recip(mark(X)) → mark(recip(X))
recip(ok(X)) → ok(recip(X))
sqr(mark(X)) → mark(sqr(X))
sqr(ok(X)) → ok(sqr(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
add(mark(X1), X2) → mark(add(X1, X2))
add(X1, mark(X2)) → mark(add(X1, X2))
add(ok(X1), ok(X2)) → ok(add(X1, X2))
dbl(mark(X)) → mark(dbl(X))
dbl(ok(X)) → ok(dbl(X))
first(mark(X1), X2) → mark(first(X1, X2))
first(X1, mark(X2)) → mark(first(X1, X2))
first(ok(X1), ok(X2)) → ok(first(X1, X2))
half(mark(X)) → mark(half(X))
half(ok(X)) → ok(half(X))

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

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

(87) PisEmptyProof (EQUIVALENT transformation)

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

(88) TRUE