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:

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

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

Q is empty.

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

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
PLUS2(s1(x), y) -> PLUS2(x, y)
MINUS2(minus2(x, y), z) -> PLUS2(y, z)
MINUS2(minus2(x, y), z) -> MINUS2(x, plus2(y, z))
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))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))

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(s1(x), y) -> PLUS2(x, y)
MINUS2(minus2(x, y), z) -> PLUS2(y, z)
MINUS2(minus2(x, y), z) -> MINUS2(x, plus2(y, z))
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))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

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

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

PLUS2(s1(x), y) -> PLUS2(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))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

PLUS2(s1(x), y) -> PLUS2(x, y)
Used argument filtering: PLUS2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ 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))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))

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
            ↳ QDPAfsSolverProof
          ↳ QDP

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

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(minus2(x, y), z) -> MINUS2(x, plus2(y, z))
Used argument filtering: MINUS2(x1, x2)  =  x1
minus2(x1, x2)  =  minus1(x1)
s1(x1)  =  x1
plus2(x1, x2)  =  x2
0  =  0
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
          ↳ 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))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

MINUS2(s1(x), s1(y)) -> MINUS2(x, y)
Used argument filtering: MINUS2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Precedence:
trivial



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))

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
            ↳ QDPAfsSolverProof

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

QUOT2(s1(x), s1(y)) -> QUOT2(minus2(x, y), s1(y))
Used argument filtering: QUOT2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
minus2(x1, x2)  =  x1
plus2(x1, x2)  =  plus2(x1, x2)
0  =  0
Used ordering: Precedence:
plus2 > s1



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ 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))
minus2(minus2(x, y), z) -> minus2(x, plus2(y, z))

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.