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:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

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:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))


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

*12(s1(x), y) -> +12(*2(x, y), y)
+12(x, s1(y)) -> +12(x, y)
*12(s1(x), y) -> *12(x, y)
FACT1(s1(x)) -> P1(s1(x))
FACT1(s1(x)) -> *12(s1(x), fact1(p1(s1(x))))
FACT1(s1(x)) -> FACT1(p1(s1(x)))

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))

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:

*12(s1(x), y) -> +12(*2(x, y), y)
+12(x, s1(y)) -> +12(x, y)
*12(s1(x), y) -> *12(x, y)
FACT1(s1(x)) -> P1(s1(x))
FACT1(s1(x)) -> *12(s1(x), fact1(p1(s1(x))))
FACT1(s1(x)) -> FACT1(p1(s1(x)))

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 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:

+12(x, s1(y)) -> +12(x, y)

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))

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.

+12(x, s1(y)) -> +12(x, y)
Used argument filtering: +12(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi 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:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))

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:

*12(s1(x), y) -> *12(x, y)

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))

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.

*12(s1(x), y) -> *12(x, y)
Used argument filtering: *12(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi 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:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))

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(s1(x)) -> FACT1(p1(s1(x)))

The TRS R consists of the following rules:

p1(s1(x)) -> x
fact1(0) -> s1(0)
fact1(s1(x)) -> *2(s1(x), fact1(p1(s1(x))))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))

The set Q consists of the following terms:

p1(s1(x0))
fact1(0)
fact1(s1(x0))
*2(0, x0)
*2(s1(x0), x1)
+2(x0, 0)
+2(x0, s1(x1))

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