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(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, X2)
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(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, X2)
activate1(X) -> X

Q is empty.

Q DP problem:
The TRS P consists of the following rules:

FACT1(X) -> ZERO1(X)
FACT1(X) -> IF3(zero1(X), n__s1(0), n__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)
ACTIVATE1(n__s1(X)) -> S1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
FACT1(X) -> FACT1(p1(X))
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__prod2(X1, X2)) -> PROD2(X1, X2)
PROD2(s1(X), Y) -> PROD2(X, Y)
ADD2(s1(X), Y) -> S1(add2(X, Y))

The TRS R consists of the following rules:

fact1(X) -> if3(zero1(X), n__s1(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, 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:

FACT1(X) -> ZERO1(X)
FACT1(X) -> IF3(zero1(X), n__s1(0), n__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)
ACTIVATE1(n__s1(X)) -> S1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
FACT1(X) -> FACT1(p1(X))
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__prod2(X1, X2)) -> PROD2(X1, X2)
PROD2(s1(X), Y) -> PROD2(X, Y)
ADD2(s1(X), Y) -> S1(add2(X, Y))

The TRS R consists of the following rules:

fact1(X) -> if3(zero1(X), n__s1(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, 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 9 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(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, 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:

fact1(X) -> if3(zero1(X), n__s1(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, 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:

PROD2(s1(X), Y) -> PROD2(X, Y)

The TRS R consists of the following rules:

fact1(X) -> if3(zero1(X), n__s1(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, 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.

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(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, 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

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), n__s1(0), n__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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
zero1(0) -> true
zero1(s1(X)) -> false
p1(s1(X)) -> X
s1(X) -> n__s1(X)
prod2(X1, X2) -> n__prod2(X1, X2)
activate1(n__s1(X)) -> s1(X)
activate1(n__prod2(X1, X2)) -> prod2(X1, X2)
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.