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.

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(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 [13,14,18] contains 1 SCC with 3 less nodes.

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

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


+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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 3 + 3·x1 + 2·x1·x2 + 3·x2   
POL(+12(x1, x2)) = 2·x2   
POL(0) = 0   
POL(1) = 0   
POL(minus1(x1)) = 3 + 3·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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.