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(minus1(x)) -> x
minux1(+2(x, y)) -> +2(minus1(y), minus1(x))
+2(minus1(x), +2(x, y)) -> y
+2(+2(x, y), minus1(y)) -> x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus1(minus1(x)) -> x
minux1(+2(x, y)) -> +2(minus1(y), minus1(x))
+2(minus1(x), +2(x, y)) -> y
+2(+2(x, y), minus1(y)) -> 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:

MINUX1(+2(x, y)) -> MINUS1(x)
MINUX1(+2(x, y)) -> +12(minus1(y), minus1(x))
MINUX1(+2(x, y)) -> MINUS1(y)

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minux1(+2(x, y)) -> +2(minus1(y), minus1(x))
+2(minus1(x), +2(x, y)) -> y
+2(+2(x, y), minus1(y)) -> 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:

MINUX1(+2(x, y)) -> MINUS1(x)
MINUX1(+2(x, y)) -> +12(minus1(y), minus1(x))
MINUX1(+2(x, y)) -> MINUS1(y)

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minux1(+2(x, y)) -> +2(minus1(y), minus1(x))
+2(minus1(x), +2(x, y)) -> y
+2(+2(x, y), minus1(y)) -> x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 3 less nodes.