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:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
SEL2(s1(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__g1(X)) -> G1(activate1(X))
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__g1(X)) -> ACTIVATE1(X)
G1(s1(X)) -> G1(X)
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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(X), cons2(Y, Z)) -> ACTIVATE1(Z)
ACTIVATE1(n__g1(X)) -> G1(activate1(X))
SEL2(s1(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
ACTIVATE1(n__g1(X)) -> ACTIVATE1(X)
G1(s1(X)) -> G1(X)
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(s1(X)) -> G1(X)
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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.
G1(s1(X)) -> G1(X)
Used argument filtering: G1(x1) = 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:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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__g1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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__f1(X)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1) = x1
n__g1(x1) = x1
n__f1(x1) = n__f1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__g1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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__g1(X)) -> ACTIVATE1(X)
Used argument filtering: ACTIVATE1(x1) = x1
n__g1(x1) = n__g1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
The TRS R consists of the following rules:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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(X), cons2(Y, Z)) -> SEL2(X, activate1(Z))
Used argument filtering: SEL2(x1, x2) = x1
s1(x1) = s1(x1)
activate1(x1) = x1
n__f1(x1) = n__f
f1(x1) = f
n__g1(x1) = n__g1(x1)
g1(x1) = g1(x1)
0 = 0
cons2(x1, x2) = cons
Used ordering: Quasi Precedence:
[n__f, f] > cons
[n__g_1, g_1] > s_1
↳ 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:
f1(X) -> cons2(X, n__f1(n__g1(X)))
g1(0) -> s1(0)
g1(s1(X)) -> s1(s1(g1(X)))
sel2(0, cons2(X, Y)) -> X
sel2(s1(X), cons2(Y, Z)) -> sel2(X, activate1(Z))
f1(X) -> n__f1(X)
g1(X) -> n__g1(X)
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__g1(X)) -> g1(activate1(X))
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.