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(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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:

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
PLUS2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> PLUS2(minus2(y, s1(s1(z))), minus2(x, s1(0)))
PLUS2(s1(x), y) -> PLUS2(x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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:

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
PLUS2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> PLUS2(minus2(y, s1(s1(z))), minus2(x, s1(0)))
PLUS2(s1(x), y) -> PLUS2(x, y)
MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
QUOT2(s1(x), s1(y)) -> MINUS2(x, y)

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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 3 SCCs with 1 less node.

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

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

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

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(x, s1(0)))

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

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

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

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

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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 strictly oriented and are deleted.


MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS2(x1, x2)  =  MINUS1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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
          ↳ QDP
            ↳ QDPOrderProof

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

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))

The TRS R consists of the following rules:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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 strictly oriented and are deleted.


QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
QUOT2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
0  =  0

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented:

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



↳ 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:

minus2(x, 0) -> x
minus2(s1(x), s1(y)) -> minus2(x, y)
quot2(0, s1(y)) -> 0
quot2(s1(x), s1(y)) -> s1(quot2(minus2(x, y), s1(y)))
plus2(0, y) -> y
plus2(s1(x), y) -> s1(plus2(x, y))
plus2(minus2(x, s1(0)), minus2(y, s1(s1(z)))) -> plus2(minus2(y, s1(s1(z))), minus2(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.