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:

minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)

Q is empty.

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

+12(minus1(+2(x, 1)), 1) -> MINUS1(x)
+12(x, minus1(y)) -> MINUS1(+2(minus1(x), y))
+12(x, minus1(y)) -> MINUS1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, minus1(y)) -> +12(minus1(x), y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)

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(minus1(+2(x, 1)), 1) -> MINUS1(x)
+12(x, minus1(y)) -> MINUS1(+2(minus1(x), y))
+12(x, minus1(y)) -> MINUS1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, minus1(y)) -> +12(minus1(x), y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

+12(x, +2(y, z)) -> +12(x, y)
+12(x, minus1(y)) -> +12(minus1(x), y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)

The TRS R consists of the following rules:

minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)

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.

+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
Used argument filtering: +12(x1, x2)  =  x2
+2(x1, x2)  =  +2(x1, x2)
minus1(x1)  =  x1
Used ordering: Quasi Precedence: trivial


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

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

+12(x, minus1(y)) -> +12(minus1(x), y)

The TRS R consists of the following rules:

minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)

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.

+12(x, minus1(y)) -> +12(minus1(x), y)
Used argument filtering: +12(x1, x2)  =  x2
minus1(x1)  =  minus1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ QDPAfsSolverProof
QDP
                  ↳ PisEmptyProof

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

minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)

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.