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:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

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(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)


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(s1(x), y) -> +12(*2(x, y), y)
*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
MINUS1(p1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> MINUS1(x)
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)
*12(p1(x), y) -> MINUS1(y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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(s1(x), y) -> +12(*2(x, y), y)
*12(s1(x), y) -> *12(x, y)
*12(p1(x), y) -> *12(x, y)
*12(p1(x), y) -> +12(*2(x, y), minus1(y))
MINUS1(p1(x)) -> MINUS1(x)
MINUS1(s1(x)) -> MINUS1(x)
+12(s1(x), y) -> +12(x, y)
+12(p1(x), y) -> +12(x, y)
*12(p1(x), y) -> MINUS1(y)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

MINUS1(s1(x)) -> MINUS1(x)
MINUS1(p1(x)) -> MINUS1(x)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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.


MINUS1(s1(x)) -> MINUS1(x)
The remaining pairs can at least by weakly be oriented.

MINUS1(p1(x)) -> MINUS1(x)
Used ordering: Combined order from the following AFS and order.
MINUS1(x1)  =  MINUS1(x1)
s1(x1)  =  s1(x1)
p1(x1)  =  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
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ QDPOrderProof
              ↳ QDP
              ↳ QDP

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

MINUS1(p1(x)) -> MINUS1(x)

The TRS R consists of the following rules:

+2(0, y) -> y
+2(s1(x), y) -> s1(+2(x, y))
+2(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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.


MINUS1(p1(x)) -> MINUS1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS1(x1)  =  MINUS1(x1)
p1(x1)  =  p1(x1)

Lexicographic Path Order [19].
Precedence:
p1 > MINUS1

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ 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(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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

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

+12(s1(x), y) -> +12(x, y)
+12(p1(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(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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)
+12(p1(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)
p1(x1)  =  p1(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
                ↳ 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(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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

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

*12(s1(x), y) -> *12(x, y)
*12(p1(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(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

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)
*12(p1(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)
p1(x1)  =  p1(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

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(p1(x), y) -> p1(+2(x, y))
minus1(0) -> 0
minus1(s1(x)) -> p1(minus1(x))
minus1(p1(x)) -> s1(minus1(x))
*2(0, y) -> 0
*2(s1(x), y) -> +2(*2(x, y), y)
*2(p1(x), y) -> +2(*2(x, y), minus1(y))

The set Q consists of the following terms:

+2(0, x0)
+2(s1(x0), x1)
+2(p1(x0), x1)
minus1(0)
minus1(s1(x0))
minus1(p1(x0))
*2(0, x0)
*2(s1(x0), x1)
*2(p1(x0), x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.