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:

minus(minus(x)) → x
minux(+(x, y)) → +(minus(y), minus(x))
+(minus(x), +(x, y)) → y
+(+(x, y), minus(y)) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

minus(minus(x)) → x
minux(+(x, y)) → +(minus(y), minus(x))
+(minus(x), +(x, y)) → y
+(+(x, y), minus(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:

MINUX(+(x, y)) → MINUS(x)
MINUX(+(x, y)) → +1(minus(y), minus(x))
MINUX(+(x, y)) → MINUS(y)

The TRS R consists of the following rules:

minus(minus(x)) → x
minux(+(x, y)) → +(minus(y), minus(x))
+(minus(x), +(x, y)) → y
+(+(x, y), minus(y)) → x

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

MINUX(+(x, y)) → MINUS(x)
MINUX(+(x, y)) → +1(minus(y), minus(x))
MINUX(+(x, y)) → MINUS(y)

The TRS R consists of the following rules:

minus(minus(x)) → x
minux(+(x, y)) → +(minus(y), minus(x))
+(minus(x), +(x, y)) → y
+(+(x, y), minus(y)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

MINUX(+(x, y)) → MINUS(x)
MINUX(+(x, y)) → +1(minus(y), minus(x))
MINUX(+(x, y)) → MINUS(y)

The TRS R consists of the following rules:

minus(minus(x)) → x
minux(+(x, y)) → +(minus(y), minus(x))
+(minus(x), +(x, y)) → y
+(+(x, y), minus(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.