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:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), s1(0))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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:

DIV2(plus2(x, y), z) -> PLUS2(div2(x, z), div2(y, z))
DIV2(s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(s1(x), s1(y)) -> P1(s1(x))
MINUS2(s1(x), s1(y)) -> P1(s1(y))
MINUS2(x, plus2(y, z)) -> MINUS2(minus2(x, y), z)
PLUS2(s1(x), y) -> PLUS2(y, minus2(s1(x), s1(0)))
MINUS2(s1(x), s1(y)) -> MINUS2(p1(s1(x)), p1(s1(y)))
P1(s1(s1(x))) -> P1(s1(x))
DIV2(s1(x), s1(y)) -> DIV2(minus2(x, y), s1(y))
MINUS2(x, plus2(y, z)) -> MINUS2(x, y)
PLUS2(s1(x), y) -> MINUS2(s1(x), s1(0))
DIV2(plus2(x, y), z) -> DIV2(x, z)
DIV2(plus2(x, y), z) -> DIV2(y, z)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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:

DIV2(plus2(x, y), z) -> PLUS2(div2(x, z), div2(y, z))
DIV2(s1(x), s1(y)) -> MINUS2(x, y)
MINUS2(s1(x), s1(y)) -> P1(s1(x))
MINUS2(s1(x), s1(y)) -> P1(s1(y))
MINUS2(x, plus2(y, z)) -> MINUS2(minus2(x, y), z)
PLUS2(s1(x), y) -> PLUS2(y, minus2(s1(x), s1(0)))
MINUS2(s1(x), s1(y)) -> MINUS2(p1(s1(x)), p1(s1(y)))
P1(s1(s1(x))) -> P1(s1(x))
DIV2(s1(x), s1(y)) -> DIV2(minus2(x, y), s1(y))
MINUS2(x, plus2(y, z)) -> MINUS2(x, y)
PLUS2(s1(x), y) -> MINUS2(s1(x), s1(0))
DIV2(plus2(x, y), z) -> DIV2(x, z)
DIV2(plus2(x, y), z) -> DIV2(y, z)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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 5 less nodes.

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

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

P1(s1(s1(x))) -> P1(s1(x))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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.


P1(s1(s1(x))) -> P1(s1(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(P1(x1)) = 2·x1   
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:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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:

MINUS2(s1(x), s1(y)) -> MINUS2(p1(s1(x)), p1(s1(y)))
MINUS2(x, plus2(y, z)) -> MINUS2(x, y)
MINUS2(x, plus2(y, z)) -> MINUS2(minus2(x, y), z)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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.


MINUS2(x, plus2(y, z)) -> MINUS2(x, y)
MINUS2(x, plus2(y, z)) -> MINUS2(minus2(x, y), z)
The remaining pairs can at least be oriented weakly.

MINUS2(s1(x), s1(y)) -> MINUS2(p1(s1(x)), p1(s1(y)))
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(MINUS2(x1, x2)) = 2·x2   
POL(minus2(x1, x2)) = 0   
POL(p1(x1)) = 0   
POL(plus2(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(s1(x1)) = 0   

The following usable rules [14] were oriented:

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



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

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

MINUS2(s1(x), s1(y)) -> MINUS2(p1(s1(x)), p1(s1(y)))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), s1(0))))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

PLUS2(s1(x), y) -> PLUS2(y, minus2(s1(x), s1(0)))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), s1(0))))

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

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

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

DIV2(s1(x), s1(y)) -> DIV2(minus2(x, y), s1(y))
DIV2(plus2(x, y), z) -> DIV2(x, z)
DIV2(plus2(x, y), z) -> DIV2(y, z)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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.


DIV2(s1(x), s1(y)) -> DIV2(minus2(x, y), s1(y))
The remaining pairs can at least be oriented weakly.

DIV2(plus2(x, y), z) -> DIV2(x, z)
DIV2(plus2(x, y), z) -> DIV2(y, z)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(DIV2(x1, x2)) = 2·x1   
POL(minus2(x1, x2)) = x1   
POL(p1(x1)) = x1   
POL(plus2(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + 2·x1   

The following usable rules [14] were oriented:

minus2(x, 0) -> x
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(0, y) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))



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

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

DIV2(plus2(x, y), z) -> DIV2(x, z)
DIV2(plus2(x, y), z) -> DIV2(y, z)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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.


DIV2(plus2(x, y), z) -> DIV2(x, z)
DIV2(plus2(x, y), z) -> DIV2(y, z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DIV2(x1, x2)) = 2·x1   
POL(plus2(x1, x2)) = 1 + 2·x1 + 2·x2   

The following usable rules [14] were oriented: none



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

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

minus2(x, 0) -> x
minus2(0, y) -> 0
minus2(s1(x), s1(y)) -> minus2(p1(s1(x)), p1(s1(y)))
minus2(x, plus2(y, z)) -> minus2(minus2(x, y), z)
p1(s1(s1(x))) -> s1(p1(s1(x)))
p1(0) -> s1(s1(0))
div2(s1(x), s1(y)) -> s1(div2(minus2(x, y), s1(y)))
div2(plus2(x, y), z) -> plus2(div2(x, z), div2(y, z))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(y, minus2(s1(x), 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.