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(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(s1(0)))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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(x, *2(x, y))
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F1(s1(x)) -> +12(*2(s1(x), s1(s1(x))), s1(s1(0)))
+12(s1(x), y) -> +12(x, y)
F1(s1(x)) -> *12(s1(x), s1(s1(x)))
F1(s1(x)) -> -12(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(s1(0))))
F1(s1(x)) -> F1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(s1(0)))))
TWICE1(s1(x)) -> TWICE1(x)
F1(s1(x)) -> *12(s1(s1(x)), s1(s1(x)))

The TRS R consists of the following rules:

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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(x, *2(x, y))
*12(x, s1(y)) -> *12(x, y)
-12(s1(x), s1(y)) -> -12(x, y)
F1(s1(x)) -> +12(*2(s1(x), s1(s1(x))), s1(s1(0)))
+12(s1(x), y) -> +12(x, y)
F1(s1(x)) -> *12(s1(x), s1(s1(x)))
F1(s1(x)) -> -12(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(s1(0))))
F1(s1(x)) -> F1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(s1(0)))))
TWICE1(s1(x)) -> TWICE1(x)
F1(s1(x)) -> *12(s1(s1(x)), s1(s1(x)))

The TRS R consists of the following rules:

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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 5 SCCs with 5 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ 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(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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(s1(x), s1(y)) -> -12(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( -12(x1, x2) ) = x1



The following usable rules [14] were oriented: none



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

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

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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
          ↳ QDP

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

TWICE1(s1(x)) -> TWICE1(x)

The TRS R consists of the following rules:

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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.


TWICE1(s1(x)) -> TWICE1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 1


POL( TWICE1(x1) ) = x1



The following usable rules [14] were oriented: none



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

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

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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
          ↳ 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(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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(s1(x), y) -> +12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( +12(x1, x2) ) = x1 + x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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
          ↳ 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(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
*2(x, 0) -> 0
*2(x, s1(y)) -> +2(x, *2(x, y))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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 Order [17,21] with Interpretation:

POL( *12(x1, x2) ) = x1 + x2


POL( s1(x1) ) = x1 + 1



The following usable rules [14] were oriented: none



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

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

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(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
          ↳ QDP

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

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

The TRS R consists of the following rules:

+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))
twice1(0) -> 0
twice1(s1(x)) -> s1(s1(twice1(x)))
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
f1(s1(x)) -> f1(-2(*2(s1(s1(x)), s1(s1(x))), +2(*2(s1(x), s1(s1(x))), s1(s1(0)))))

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