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:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
GEQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__s1(X)) -> S1(X)
DIV2(s1(X), n__s1(Y)) -> IF3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
IF3(true, X, Y) -> ACTIVATE1(X)
GEQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(X)
MINUS2(n__s1(X), n__s1(Y)) -> ACTIVATE1(Y)
DIV2(s1(X), n__s1(Y)) -> ACTIVATE1(Y)
DIV2(s1(X), n__s1(Y)) -> GEQ2(X, activate1(Y))
MINUS2(n__s1(X), n__s1(Y)) -> MINUS2(activate1(X), activate1(Y))
MINUS2(n__s1(X), n__s1(Y)) -> ACTIVATE1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
MINUS2(n__0, Y) -> 01
DIV2(s1(X), n__s1(Y)) -> MINUS2(X, activate1(Y))
GEQ2(n__s1(X), n__s1(Y)) -> GEQ2(activate1(X), activate1(Y))
DIV2(s1(X), n__s1(Y)) -> DIV2(minus2(X, activate1(Y)), n__s1(activate1(Y)))
The TRS R consists of the following rules:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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:
GEQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(Y)
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__s1(X)) -> S1(X)
DIV2(s1(X), n__s1(Y)) -> IF3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
IF3(true, X, Y) -> ACTIVATE1(X)
GEQ2(n__s1(X), n__s1(Y)) -> ACTIVATE1(X)
MINUS2(n__s1(X), n__s1(Y)) -> ACTIVATE1(Y)
DIV2(s1(X), n__s1(Y)) -> ACTIVATE1(Y)
DIV2(s1(X), n__s1(Y)) -> GEQ2(X, activate1(Y))
MINUS2(n__s1(X), n__s1(Y)) -> MINUS2(activate1(X), activate1(Y))
MINUS2(n__s1(X), n__s1(Y)) -> ACTIVATE1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
MINUS2(n__0, Y) -> 01
DIV2(s1(X), n__s1(Y)) -> MINUS2(X, activate1(Y))
GEQ2(n__s1(X), n__s1(Y)) -> GEQ2(activate1(X), activate1(Y))
DIV2(s1(X), n__s1(Y)) -> DIV2(minus2(X, activate1(Y)), n__s1(activate1(Y)))
The TRS R consists of the following rules:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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 13 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
GEQ2(n__s1(X), n__s1(Y)) -> GEQ2(activate1(X), activate1(Y))
The TRS R consists of the following rules:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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.
GEQ2(n__s1(X), n__s1(Y)) -> GEQ2(activate1(X), activate1(Y))
Used argument filtering: GEQ2(x1, x2) = x2
n__s1(x1) = n__s1(x1)
activate1(x1) = x1
n__0 = n__0
0 = 0
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
[n__s_1, s_1]
[n__0, 0]
↳ 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:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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:
MINUS2(n__s1(X), n__s1(Y)) -> MINUS2(activate1(X), activate1(Y))
The TRS R consists of the following rules:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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.
MINUS2(n__s1(X), n__s1(Y)) -> MINUS2(activate1(X), activate1(Y))
Used argument filtering: MINUS2(x1, x2) = x2
n__s1(x1) = n__s1(x1)
activate1(x1) = x1
n__0 = n__0
0 = 0
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
[n__s_1, s_1]
[n__0, 0]
↳ 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:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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:
DIV2(s1(X), n__s1(Y)) -> DIV2(minus2(X, activate1(Y)), n__s1(activate1(Y)))
The TRS R consists of the following rules:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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.
DIV2(s1(X), n__s1(Y)) -> DIV2(minus2(X, activate1(Y)), n__s1(activate1(Y)))
Used argument filtering: DIV2(x1, x2) = x1
s1(x1) = s
minus2(x1, x2) = minus
activate1(x1) = x1
n__0 = n__0
0 = 0
n__s1(x1) = n__s
Used ordering: Quasi Precedence:
[s, n__s] > [minus, n__0, 0]
↳ 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:
minus2(n__0, Y) -> 0
minus2(n__s1(X), n__s1(Y)) -> minus2(activate1(X), activate1(Y))
geq2(X, n__0) -> true
geq2(n__0, n__s1(Y)) -> false
geq2(n__s1(X), n__s1(Y)) -> geq2(activate1(X), activate1(Y))
div2(0, n__s1(Y)) -> 0
div2(s1(X), n__s1(Y)) -> if3(geq2(X, activate1(Y)), n__s1(div2(minus2(X, activate1(Y)), n__s1(activate1(Y)))), n__0)
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(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.