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 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 DependencyGraphProof (⇔)
    ↳24 TRUE

(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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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)
SQR(s(X)) → ADD(sqr(X), dbl(X))
SQR(s(X)) → SQR(X)
SQR(s(X)) → DBL(X)
DBL(s(X)) → DBL(X)
ADD(s(X), Y) → ADD(X, Y)
FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__terms(X)) → TERMS(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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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 4 SCCs with 4 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
recip1 > s1
sqr1 > [0, first1, nil, activate1] > terms1 > cons > s1
sqr1 > [0, first1, nil, activate1] > terms1 > nterms1 > s1
sqr1 > add2 > s1
sqr1 > dbl1 > s1

Status:
s1: [1]
terms1: [1]
cons: []
recip1: [1]
sqr1: [1]
nterms1: [1]
0: []
add2: [1,2]
dbl1: [1]
first1: [1]
nil: []
activate1: [1]


The following usable rules [FROCOS05] were oriented:

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

(7) 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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

DBL(s(X)) → DBL(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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[terms, sqr1, nterms, dbl1] > recip1 > cons
[terms, sqr1, nterms, dbl1] > [0, nil] > cons
[terms, sqr1, nterms, dbl1] > add2 > s1 > cons
[first, nfirst] > [0, nil] > cons

Status:
s1: [1]
terms: []
cons: []
recip1: [1]
sqr1: [1]
nterms: []
0: []
add2: [1,2]
dbl1: [1]
first: []
nil: []
nfirst: []


The following usable rules [FROCOS05] were oriented:

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

(12) 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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

SQR(s(X)) → SQR(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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[terms, sqr1, nterms, dbl1] > recip1 > cons
[terms, sqr1, nterms, dbl1] > [0, nil] > cons
[terms, sqr1, nterms, dbl1] > add2 > s1 > cons
[first, nfirst] > [0, nil] > cons

Status:
s1: [1]
terms: []
cons: []
recip1: [1]
sqr1: [1]
nterms: []
0: []
add2: [1,2]
dbl1: [1]
first: []
nil: []
nfirst: []


The following usable rules [FROCOS05] were oriented:

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

(17) 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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) Obligation:

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

ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__first(x1, x2)  =  x2
FIRST(x1, x2)  =  FIRST(x2)
s(x1)  =  s
cons(x1, x2)  =  cons(x1, x2)
terms(x1)  =  terms(x1)
recip(x1)  =  recip(x1)
sqr(x1)  =  sqr
n__terms(x1)  =  n__terms(x1)
0  =  0
add(x1, x2)  =  x2
dbl(x1)  =  x1
first(x1, x2)  =  first(x2)
nil  =  nil
activate(x1)  =  activate(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[ACTIVATE1, FIRST1] > [s, recip1, sqr, nterms1, 0]
[first1, activate1] > terms1 > cons2 > [s, recip1, sqr, nterms1, 0]
[first1, activate1] > nil > [s, recip1, sqr, nterms1, 0]

Status:
ACTIVATE1: [1]
FIRST1: [1]
s: []
cons2: [1,2]
terms1: [1]
recip1: [1]
sqr: []
nterms1: [1]
0: []
first1: [1]
nil: []
activate1: [1]


The following usable rules [FROCOS05] were oriented:

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

(22) Obligation:

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

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(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__terms(X)) → terms(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.

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(24) TRUE