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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 3 SCCs with 7 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), 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.
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 Order [17,21] with Interpretation:

POL( ADD2(x1, x2) ) = max{0, x1 - 2}


POL( s1(x1) ) = x1 + 3



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), 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
            ↳ 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), 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.
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 Order [17,21] with Interpretation:

POL( PROD2(x1, x2) ) = max{0, x1 - 2}


POL( s1(x1) ) = x1 + 3



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), 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.