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:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> s1(0)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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:

*12(x, s1(y)) -> +12(*2(x, y), x)
FLOOP2(s1(x), y) -> *12(s1(x), y)
FAC1(s1(x)) -> FAC1(x)
+12(x, s1(y)) -> +12(x, y)
FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))
FAC1(s1(x)) -> *12(s1(x), fac1(x))
*12(x, s1(y)) -> *12(x, y)
FAC1(0) -> 11

The TRS R consists of the following rules:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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:

*12(x, s1(y)) -> +12(*2(x, y), x)
FLOOP2(s1(x), y) -> *12(s1(x), y)
FAC1(s1(x)) -> FAC1(x)
+12(x, s1(y)) -> +12(x, y)
FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))
FAC1(s1(x)) -> *12(s1(x), fac1(x))
*12(x, s1(y)) -> *12(x, y)
FAC1(0) -> 11

The TRS R consists of the following rules:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ 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:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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.


+12(x, s1(y)) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+12(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

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:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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:

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

The TRS R consists of the following rules:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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.


*12(x, s1(y)) -> *12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*12(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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:

FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))

The TRS R consists of the following rules:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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.


FLOOP2(s1(x), y) -> FLOOP2(x, *2(s1(x), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(*2(x1, x2)) = 0   
POL(+2(x1, x2)) = 0   
POL(0) = 0   
POL(FLOOP2(x1, x2)) = 2·x1   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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:

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

The TRS R consists of the following rules:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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.


FAC1(s1(x)) -> FAC1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(FAC1(x1)) = 2·x1   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ 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:

fac1(0) -> 1
fac1(s1(x)) -> *2(s1(x), fac1(x))
floop2(0, y) -> y
floop2(s1(x), y) -> floop2(x, *2(s1(x), y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(*2(x, y), x)
+2(x, 0) -> x
+2(x, s1(y)) -> s1(+2(x, y))
1 -> s1(0)
fac1(0) -> 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.