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
  ↳ 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

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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

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) -> 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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ 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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ADD2(s1(X), Y) -> ADD2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ADD2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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

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
            ↳ QDPOrderProof
          ↳ 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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROD2(s1(X), Y) -> PROD2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROD2(x1, x2)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ 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

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