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:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FACT(X) → PROD(X, fact(p(X)))
PROD(s(X), Y) → ADD(Y, prod(X, Y))
FACT(X) → P(X)
ADD(s(X), Y) → ADD(X, Y)
FACT(X) → FACT(p(X))
PROD(s(X), Y) → PROD(X, Y)
FACT(X) → ZERO(X)
FACT(X) → IF(zero(X), s(0), prod(X, fact(p(X))))

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(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:

FACT(X) → PROD(X, fact(p(X)))
PROD(s(X), Y) → ADD(Y, prod(X, Y))
FACT(X) → P(X)
ADD(s(X), Y) → ADD(X, Y)
FACT(X) → FACT(p(X))
PROD(s(X), Y) → PROD(X, Y)
FACT(X) → ZERO(X)
FACT(X) → IF(zero(X), s(0), prod(X, fact(p(X))))

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


ADD(s(X), Y) → ADD(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 4 + (4)x_1   
POL(ADD(x1, x2)) = (2)x_1   
The value of delta used in the strict ordering is 8.
The following usable rules [17] 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:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(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:

PROD(s(X), Y) → PROD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


PROD(s(X), Y) → PROD(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1/4 + (2)x_1   
POL(PROD(x1, x2)) = (1/4)x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] 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:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(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:

FACT(X) → FACT(p(X))

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

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