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:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))

Q is empty.

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

TIMES2(s1(x), y) -> PLUS2(times2(x, y), y)
P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
TIMES2(s1(x), y) -> TIMES2(x, y)
FAC1(s1(x)) -> TIMES2(fac1(p1(s1(x))), s1(x))
PLUS2(x, s1(y)) -> PLUS2(x, y)
FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(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:

TIMES2(s1(x), y) -> PLUS2(times2(x, y), y)
P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
TIMES2(s1(x), y) -> TIMES2(x, y)
FAC1(s1(x)) -> TIMES2(fac1(p1(s1(x))), s1(x))
PLUS2(x, s1(y)) -> PLUS2(x, y)
FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 4 SCCs with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPAfsSolverProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

P1(s1(s1(x))) -> P1(s1(x))

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(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.

P1(s1(s1(x))) -> P1(s1(x))
Used argument filtering: P1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(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
          ↳ QDP

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

PLUS2(x, s1(y)) -> PLUS2(x, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(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.

PLUS2(x, s1(y)) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(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
            ↳ QDPAfsSolverProof
          ↳ QDP

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

TIMES2(s1(x), y) -> TIMES2(x, y)

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(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.

TIMES2(s1(x), y) -> TIMES2(x, y)
Used argument filtering: TIMES2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(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
QDP

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

FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

plus2(x, 0) -> x
plus2(x, s1(y)) -> s1(plus2(x, y))
times2(0, y) -> 0
times2(x, 0) -> 0
times2(s1(x), y) -> plus2(times2(x, y), y)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0
fac1(s1(x)) -> times2(fac1(p1(s1(x))), s1(x))

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