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:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(activate1(X))
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(activate1(X))
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__fact1(X)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ADD2(X, Y)
PROD2(s1(X), Y) -> ADD2(Y, prod2(X, Y))
ACTIVATE1(n__fact1(X)) -> FACT1(activate1(X))
ACTIVATE1(n__0) -> 01
IF3(true, X, Y) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> S1(add2(X, Y))
ACTIVATE1(n__prod2(X1, X2)) -> PROD2(activate1(X1), activate1(X2))
ACTIVATE1(n__prod2(X1, X2)) -> ACTIVATE1(X2)
FACT1(X) -> ZERO1(X)
ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__prod2(X1, X2)) -> ACTIVATE1(X1)
IF3(false, X, Y) -> ACTIVATE1(Y)
FACT1(X) -> IF3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
ACTIVATE1(n__p1(X)) -> P1(activate1(X))
PROD2(s1(X), Y) -> PROD2(X, Y)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(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:
ACTIVATE1(n__fact1(X)) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> ADD2(X, Y)
PROD2(s1(X), Y) -> ADD2(Y, prod2(X, Y))
ACTIVATE1(n__fact1(X)) -> FACT1(activate1(X))
ACTIVATE1(n__0) -> 01
IF3(true, X, Y) -> ACTIVATE1(X)
ADD2(s1(X), Y) -> S1(add2(X, Y))
ACTIVATE1(n__prod2(X1, X2)) -> PROD2(activate1(X1), activate1(X2))
ACTIVATE1(n__prod2(X1, X2)) -> ACTIVATE1(X2)
FACT1(X) -> ZERO1(X)
ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__prod2(X1, X2)) -> ACTIVATE1(X1)
IF3(false, X, Y) -> ACTIVATE1(Y)
FACT1(X) -> IF3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
ACTIVATE1(n__p1(X)) -> P1(activate1(X))
PROD2(s1(X), Y) -> PROD2(X, Y)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(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 7 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:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(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.
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:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(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:
PROD2(s1(X), Y) -> PROD2(X, Y)
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(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.
PROD2(s1(X), Y) -> PROD2(X, Y)
Used argument filtering: PROD2(x1, x2) = x1
s1(x1) = s1(x1)
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:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(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
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__fact1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__prod2(X1, X2)) -> ACTIVATE1(X2)
ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__prod2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__fact1(X)) -> FACT1(activate1(X))
IF3(false, X, Y) -> ACTIVATE1(Y)
FACT1(X) -> IF3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), n__s1(n__0), n__prod2(X, n__fact1(n__p1(X))))
add2(0, X) -> X
add2(s1(X), Y) -> s1(add2(X, Y))
prod2(0, X) -> 0
prod2(s1(X), Y) -> add2(Y, prod2(X, Y))
if3(true, X, Y) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
0 -> n__0
prod2(X1, X2) -> n__prod2(X1, X2)
fact1(X) -> n__fact1(X)
p1(X) -> n__p1(X)
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__0) -> 0
activate1(n__prod2(X1, X2)) -> prod2(activate1(X1), activate1(X2))
activate1(n__fact1(X)) -> fact1(activate1(X))
activate1(n__p1(X)) -> p1(activate1(X))
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.