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:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ADD2(s1(X), Y) -> ADD2(X, Y)
ACTIVATE1(n__fib12(X1, X2)) -> FIB12(activate1(X1), activate1(X2))
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
FIB1(N) -> SEL2(N, fib12(s1(0), s1(0)))
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(activate1(X1), activate1(X2))
FIB1(N) -> FIB12(s1(0), s1(0))
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(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:
SEL2(s1(N), cons2(X, XS)) -> ACTIVATE1(XS)
ADD2(s1(X), Y) -> ADD2(X, Y)
ACTIVATE1(n__fib12(X1, X2)) -> FIB12(activate1(X1), activate1(X2))
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
FIB1(N) -> SEL2(N, fib12(s1(0), s1(0)))
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__add2(X1, X2)) -> ADD2(activate1(X1), activate1(X2))
FIB1(N) -> FIB12(s1(0), s1(0))
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(X2))
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 5 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ 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:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(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
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(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.
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__fib12(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__add2(X1, X2)) -> ACTIVATE1(X1)
Used argument filtering: ACTIVATE1(x1) = x1
n__add2(x1, x2) = n__add2(x1, x2)
n__fib12(x1, x2) = n__fib12(x1, x2)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(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
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(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.
SEL2(s1(N), cons2(X, XS)) -> SEL2(N, activate1(XS))
Used argument filtering: SEL2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
fib1(N) -> sel2(N, fib12(s1(0), s1(0)))
fib12(X, Y) -> cons2(X, n__fib12(Y, n__add2(X, Y)))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
sel2(0, cons2(X, XS)) -> X
sel2(s1(N), cons2(X, XS)) -> sel2(N, activate1(XS))
fib12(X1, X2) -> n__fib12(X1, X2)
add2(X1, X2) -> n__add2(X1, X2)
activate1(n__fib12(X1, X2)) -> fib12(activate1(X1), activate1(X2))
activate1(n__add2(X1, X2)) -> add2(activate1(X1), activate1(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.