(0) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

TERMS(N) → SQR(N)
TERMS(N) → S(N)
SQR(s(X)) → S(n__add(sqr(activate(X)), dbl(activate(X))))
SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → S(n__s(n__dbl(activate(X))))
DBL(s(X)) → ACTIVATE(X)
ADD(s(X), Y) → S(n__add(activate(X), Y))
ADD(s(X), Y) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__terms(X)) → TERMS(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__dbl(X)) → DBL(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 5 less nodes.

(4) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__dbl(X)) → DBL(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__dbl(X)) → DBL(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(SQR(x1)) = -I + 0A·x1

POL(s(x1)) = 0A + 0A·x1

POL(activate(x1)) = -I + 0A·x1

POL(ACTIVATE(x1)) = -I + 0A·x1

POL(n__terms(x1)) = 0A + 0A·x1

POL(TERMS(x1)) = 0A + 0A·x1

POL(DBL(x1)) = -I + 0A·x1

POL(n__add(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(ADD(x1, x2)) = 0A + 0A·x1 + -I·x2

POL(n__dbl(x1)) = 0A + 1A·x1

POL(n__first(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(FIRST(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(cons(x1, x2)) = 0A + -I·x1 + 0A·x2

POL(dbl(x1)) = 0A + 1A·x1

POL(0) = 1A

POL(sqr(x1)) = 0A + 1A·x1

POL(add(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(n__s(x1)) = 0A + 0A·x1

POL(terms(x1)) = 0A + 0A·x1

POL(recip(x1)) = -I + 0A·x1

POL(first(x1, x2)) = 0A + 0A·x1 + 0A·x2

POL(nil) = 0A

The following usable rules [FROCOS05] were oriented:

dbl(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
add(0, X) → X
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
sqr(0) → 0
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

(6) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
ACTIVATE(n__terms(X)) → TERMS(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVATE(n__terms(X)) → TERMS(X)
ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(ADD(x1, x2)) = x1 + x2   
POL(DBL(x1)) = x1   
POL(FIRST(x1, x2)) = x1 + x2   
POL(SQR(x1)) = x1   
POL(TERMS(x1)) = x1   
POL(activate(x1)) = x1   
POL(add(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = x2   
POL(dbl(x1)) = 0   
POL(first(x1, x2)) = 1 + x1 + x2   
POL(n__add(x1, x2)) = x1 + x2   
POL(n__dbl(x1)) = 0   
POL(n__first(x1, x2)) = 1 + x1 + x2   
POL(n__s(x1)) = x1   
POL(n__terms(x1)) = 1 + x1   
POL(nil) = 0   
POL(recip(x1)) = 0   
POL(s(x1)) = x1   
POL(sqr(x1)) = 0   
POL(terms(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

dbl(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
add(0, X) → X
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
sqr(0) → 0
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

(8) Obligation:

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

SQR(s(X)) → SQR(activate(X))
SQR(s(X)) → ACTIVATE(X)
TERMS(N) → SQR(N)
SQR(s(X)) → DBL(activate(X))
DBL(s(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
ADD(s(X), Y) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(X)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 6 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

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

ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(12) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(13) Obligation:

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

ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)

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

(14) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ACTIVATE(n__add(X1, X2)) → ADD(X1, X2)
    The graph contains the following edges 1 > 1, 1 > 2

  • ADD(s(X), Y) → ACTIVATE(X)
    The graph contains the following edges 1 > 1

(15) TRUE

(16) Obligation:

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

SQR(s(X)) → SQR(activate(X))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


SQR(s(X)) → SQR(activate(X))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SQR(x1)  =  x1
s(x1)  =  s(x1)
activate(x1)  =  x1
dbl(x1)  =  dbl(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
n__s(x1)  =  n__s(x1)
n__dbl(x1)  =  n__dbl(x1)
terms(x1)  =  terms
cons(x1, x2)  =  cons
n__terms(x1)  =  n__terms
first(x1, x2)  =  first
n__first(x1, x2)  =  n__first
n__add(x1, x2)  =  n__add(x1, x2)
nil  =  nil

Recursive path order with status [RPO].
Quasi-Precedence:
sqr1 > [dbl1, ndbl1] > [s1, ns1]
sqr1 > [nadd2, add2] > [s1, ns1]
[terms, nterms] > cons
[first, nfirst] > cons
[first, nfirst] > nil

Status:
add2: [1,2]
nadd2: [1,2]


The following usable rules [FROCOS05] were oriented:

dbl(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
add(0, X) → X
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
sqr(0) → 0
terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

(18) Obligation:

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

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(n__add(sqr(activate(X)), dbl(activate(X))))
dbl(0) → 0
dbl(s(X)) → s(n__s(n__dbl(activate(X))))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(activate(X), activate(Z)))
terms(X) → n__terms(X)
add(X1, X2) → n__add(X1, X2)
s(X) → n__s(X)
dbl(X) → n__dbl(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(X)
activate(n__add(X1, X2)) → add(X1, X2)
activate(n__s(X)) → s(X)
activate(n__dbl(X)) → dbl(X)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE