Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(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, first2(X, Z))
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(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, first2(X, Z))
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
Q DP problem:
The TRS P consists of the following rules:
SQR1(s1(X)) -> ADD2(sqr1(X), dbl1(X))
TERMS1(N) -> TERMS1(s1(N))
ADD2(s1(X), Y) -> ADD2(X, Y)
SQR1(s1(X)) -> SQR1(X)
SQR1(s1(X)) -> DBL1(X)
TERMS1(N) -> SQR1(N)
DBL1(s1(X)) -> DBL1(X)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
SQR1(s1(X)) -> ADD2(sqr1(X), dbl1(X))
TERMS1(N) -> TERMS1(s1(N))
ADD2(s1(X), Y) -> ADD2(X, Y)
SQR1(s1(X)) -> SQR1(X)
SQR1(s1(X)) -> DBL1(X)
TERMS1(N) -> SQR1(N)
DBL1(s1(X)) -> DBL1(X)
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 5 SCCs with 3 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
FIRST2(s1(X), cons2(Y, Z)) -> FIRST2(X, Z)
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
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)) -> FIRST2(X, Z)
Used argument filtering: FIRST2(x1, x2) = x2
cons2(x1, x2) = cons1(x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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)), 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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ 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)), 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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
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: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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)), 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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ 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)), 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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
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: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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)), 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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ 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)), 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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
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: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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)), 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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TERMS1(N) -> TERMS1(s1(N))
The TRS R consists of the following rules:
terms1(N) -> cons2(recip1(sqr1(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, first2(X, Z))
The set Q consists of the following terms:
terms1(x0)
sqr1(0)
sqr1(s1(x0))
dbl1(0)
dbl1(s1(x0))
add2(0, x0)
add2(s1(x0), x1)
first2(0, x0)
first2(s1(x0), cons2(x1, x2))
We have to consider all minimal (P,Q,R)-chains.