Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
SQR1(s1(X)) -> ADD2(sqr1(X), dbl1(X))
ADD2(s1(X), Y) -> ADD2(X, Y)
SQR1(s1(X)) -> SQR1(X)
SQR1(s1(X)) -> DBL1(X)
TERMS1(N) -> SQR1(N)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
DBL1(s1(X)) -> DBL1(X)
ACTIVATE1(n__terms1(X)) -> TERMS1(X)
HALF1(s1(s1(X))) -> HALF1(X)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
SQR1(s1(X)) -> ADD2(sqr1(X), dbl1(X))
ADD2(s1(X), Y) -> ADD2(X, Y)
SQR1(s1(X)) -> SQR1(X)
SQR1(s1(X)) -> DBL1(X)
TERMS1(N) -> SQR1(N)
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
DBL1(s1(X)) -> DBL1(X)
ACTIVATE1(n__terms1(X)) -> TERMS1(X)
HALF1(s1(s1(X))) -> HALF1(X)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HALF1(s1(s1(X))) -> HALF1(X)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
HALF1(s1(s1(X))) -> HALF1(X)
Used argument filtering: HALF1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ADD2(s1(X), Y) -> ADD2(X, Y)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ADD2(s1(X), Y) -> ADD2(X, Y)
Used argument filtering: ADD2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
DBL1(s1(X)) -> DBL1(X)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
DBL1(s1(X)) -> DBL1(X)
Used argument filtering: DBL1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SQR1(s1(X)) -> SQR1(X)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
SQR1(s1(X)) -> SQR1(X)
Used argument filtering: SQR1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
FIRST2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
Used argument filtering: ACTIVATE1(x1) = x1
n__first2(x1, x2) = x2
FIRST2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__first2(X1, X2)) -> FIRST2(X1, X2)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(N)), n__terms1(s1(N)))
sqr1(0) -> 0
sqr1(s1(X)) -> s1(add2(sqr1(X), dbl1(X)))
dbl1(0) -> 0
dbl1(s1(X)) -> s1(s1(dbl1(X)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
first2(0, X) -> nil
first2(s1(X), cons2(Y, Z)) -> cons2(Y, n__first2(X, activate1(Z)))
half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(X))) -> s1(half1(X))
half1(dbl1(X)) -> X
terms1(X) -> n__terms1(X)
first2(X1, X2) -> n__first2(X1, X2)
activate1(n__terms1(X)) -> terms1(X)
activate1(n__first2(X1, X2)) -> first2(X1, X2)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.