(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)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
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)
HALF(s(s(X))) → HALF(X)
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)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
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 5 SCCs with 4 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(X))) → HALF(X)
The TRS R consists of the following rules:
terms(N) → cons(recip(sqr(N)), 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)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
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) 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.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
HALF(s(s(X))) → HALF(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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:
- HALF(s(s(X))) → HALF(X)
The graph contains the following edges 1 > 1
(9) TRUE
(10) 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)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
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) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ADD(s(X), Y) → ADD(X, Y)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) 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:
- ADD(s(X), Y) → ADD(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2
(14) TRUE
(15) 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)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
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) 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.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
DBL(s(X)) → DBL(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) 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:
- DBL(s(X)) → DBL(X)
The graph contains the following edges 1 > 1
(19) TRUE
(20) 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)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
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) 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.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SQR(s(X)) → SQR(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) 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:
- SQR(s(X)) → SQR(X)
The graph contains the following edges 1 > 1
(24) TRUE
(25) 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)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
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.
(26) 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.
(27) 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)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) 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:
- FIRST(s(X), cons(Y, Z)) → ACTIVATE(Z)
The graph contains the following edges 2 > 1
- ACTIVATE(n__first(X1, X2)) → FIRST(X1, X2)
The graph contains the following edges 1 > 1, 1 > 2
(29) TRUE