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), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
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:
fact1(X) -> if3(zero1(X), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
Q DP problem:
The TRS P consists of the following rules:
FACT1(X) -> ZERO1(X)
FACT1(X) -> PROD2(X, fact1(p1(X)))
PROD2(s1(X), Y) -> ADD2(Y, prod2(X, Y))
ADD2(s1(X), Y) -> ADD2(X, Y)
FACT1(X) -> P1(X)
FACT1(X) -> IF3(zero1(X), s1(0), prod2(X, fact1(p1(X))))
FACT1(X) -> FACT1(p1(X))
PROD2(s1(X), Y) -> PROD2(X, Y)
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
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:
FACT1(X) -> ZERO1(X)
FACT1(X) -> PROD2(X, fact1(p1(X)))
PROD2(s1(X), Y) -> ADD2(Y, prod2(X, Y))
ADD2(s1(X), Y) -> ADD2(X, Y)
FACT1(X) -> P1(X)
FACT1(X) -> IF3(zero1(X), s1(0), prod2(X, fact1(p1(X))))
FACT1(X) -> FACT1(p1(X))
PROD2(s1(X), Y) -> PROD2(X, Y)
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 5 less nodes.
↳ QTRS
↳ Non-Overlap Check
↳ 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), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
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
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
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
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), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
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: Precedence:
trivial
↳ QTRS
↳ Non-Overlap Check
↳ 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), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
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
Q DP problem:
The TRS P consists of the following rules:
FACT1(X) -> FACT1(p1(X))
The TRS R consists of the following rules:
fact1(X) -> if3(zero1(X), s1(0), prod2(X, fact1(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) -> X
if3(false, X, Y) -> Y
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
The set Q consists of the following terms:
fact1(x0)
add2(0, x0)
add2(s1(x0), x1)
prod2(0, x0)
prod2(s1(x0), x1)
if3(true, x0, x1)
if3(false, x0, x1)
zero1(0)
zero1(s1(x0))
p1(s1(x0))
We have to consider all minimal (P,Q,R)-chains.