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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

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:

-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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))


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)
F1(s1(x)) -> F1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))
-12(s1(x), s1(y)) -> -12(x, y)
F1(s1(x)) -> -12(*2(s1(s1(0)), s1(x)), s1(s1(x)))
+12(s1(x), y) -> +12(x, y)
F1(s1(x)) -> *12(s1(s1(0)), 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

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(x, s1(y)) -> +12(x, *2(x, y))
*12(x, s1(y)) -> *12(x, y)
F1(s1(x)) -> F1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))
-12(s1(x), s1(y)) -> -12(x, y)
F1(s1(x)) -> -12(*2(s1(s1(0)), s1(x)), s1(s1(x)))
+12(s1(x), y) -> +12(x, y)
F1(s1(x)) -> *12(s1(s1(0)), 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 3 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


+12(s1(x), y) -> +12(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
+12(x1, x2)  =  +11(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[+^11, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

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
                ↳ 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


*12(x, s1(y)) -> *12(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
*12(x1, x2)  =  *11(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[*^11, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

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
                ↳ 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


-12(s1(x), s1(y)) -> -12(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
-12(x1, x2)  =  -11(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

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
              ↳ QDP

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

F1(s1(x)) -> F1(-2(*2(s1(s1(0)), s1(x)), s1(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))
f1(s1(x)) -> f1(-2(*2(s1(s1(0)), s1(x)), s1(s1(x))))

The set Q consists of the following terms:

-2(x0, 0)
-2(s1(x0), s1(x1))
+2(0, x0)
+2(s1(x0), x1)
*2(x0, 0)
*2(x0, s1(x1))
f1(s1(x0))

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