Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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:

FAC2(s1(x), y) -> FAC2(p1(s1(x)), times2(s1(x), y))
PLUS2(s1(x), y) -> PLUS2(p1(s1(x)), y)
PLUS2(s1(x), y) -> P1(s1(x))
P1(s1(s1(x))) -> P1(s1(x))
TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)
TIMES2(s1(x), y) -> P1(s1(x))
FACTORIAL1(x) -> FAC2(x, s1(0))
FAC2(s1(x), y) -> P1(s1(x))
FAC2(s1(x), y) -> TIMES2(s1(x), y)
TIMES2(s1(x), y) -> PLUS2(y, times2(p1(s1(x)), y))

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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:

FAC2(s1(x), y) -> FAC2(p1(s1(x)), times2(s1(x), y))
PLUS2(s1(x), y) -> PLUS2(p1(s1(x)), y)
PLUS2(s1(x), y) -> P1(s1(x))
P1(s1(s1(x))) -> P1(s1(x))
TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)
TIMES2(s1(x), y) -> P1(s1(x))
FACTORIAL1(x) -> FAC2(x, s1(0))
FAC2(s1(x), y) -> P1(s1(x))
FAC2(s1(x), y) -> TIMES2(s1(x), y)
TIMES2(s1(x), y) -> PLUS2(y, times2(p1(s1(x)), y))

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ QDPOrderProof
          ↳ 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(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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.


P1(s1(s1(x))) -> P1(s1(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( P1(x1) ) = max{0, x1 - 1}


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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
          ↳ QDP

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

PLUS2(s1(x), y) -> PLUS2(p1(s1(x)), y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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.


PLUS2(s1(x), y) -> PLUS2(p1(s1(x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 1


POL( s1(x1) ) = x1 + 1


POL( PLUS2(x1, x2) ) = x1 + x2 + 1


POL( p1(x1) ) = max{0, x1 - 1}



The following usable rules [14] were oriented:

p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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
            ↳ QDPOrderProof
          ↳ QDP

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

TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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.


TIMES2(s1(x), y) -> TIMES2(p1(s1(x)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 1


POL( s1(x1) ) = x1 + 1


POL( TIMES2(x1, x2) ) = x1 + x2 + 1


POL( p1(x1) ) = max{0, x1 - 1}



The following usable rules [14] were oriented:

p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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
            ↳ QDPOrderProof

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

FAC2(s1(x), y) -> FAC2(p1(s1(x)), times2(s1(x), y))

The TRS R consists of the following rules:

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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.


FAC2(s1(x), y) -> FAC2(p1(s1(x)), times2(s1(x), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( 0 ) = 1


POL( FAC2(x1, x2) ) = x1


POL( times2(x1, x2) ) = max{0, x2 - 1}


POL( plus2(x1, x2) ) = x1 + x2 + 1


POL( p1(x1) ) = max{0, x1 - 1}



The following usable rules [14] were oriented:

p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(s1(0)) -> 0



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

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

plus2(0, x) -> x
plus2(s1(x), y) -> s1(plus2(p1(s1(x)), y))
times2(0, y) -> 0
times2(s1(x), y) -> plus2(y, times2(p1(s1(x)), y))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))
fac2(0, x) -> x
fac2(s1(x), y) -> fac2(p1(s1(x)), times2(s1(x), y))
factorial1(x) -> fac2(x, s1(0))

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.