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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

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(x, *2(x, y))
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
+12(s1(x), y) -> +12(x, y)
F1(s1(x)) -> *12(s1(x), s1(x))
F1(s1(x)) -> P1(*2(s1(x), s1(x)))
F1(s1(x)) -> -12(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x)))
F1(s1(x)) -> F1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(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:

*12(x, s1(y)) -> +12(x, *2(x, y))
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
+12(s1(x), y) -> +12(x, y)
F1(s1(x)) -> *12(s1(x), s1(x))
F1(s1(x)) -> P1(*2(s1(x), s1(x)))
F1(s1(x)) -> -12(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x)))
F1(s1(x)) -> F1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

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

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

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

POL(+12(x1, x2)) = x1   
POL(s1(x1)) = 1 + 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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(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
            ↳ 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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

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)) = x2   
POL(s1(x1)) = 1 + 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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(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
            ↳ QDPOrderProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

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(s1(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)) = x1   
POL(s1(x1)) = 1 + 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:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(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:

F1(s1(x)) -> F1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

The TRS R consists of the following rules:

-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
p1(s1(x)) -> x
f1(s1(x)) -> f1(-2(p1(*2(s1(x), s1(x))), *2(s1(x), s1(x))))

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