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
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(x))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MINUS(h(x)) → MINUS(x)
MINUS(f(x, y)) → MINUS(y)
MINUS(f(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(x))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ QDPOrderProof

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

MINUS(h(x)) → MINUS(x)
MINUS(f(x, y)) → MINUS(y)
MINUS(f(x, y)) → MINUS(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


MINUS(h(x)) → MINUS(x)
MINUS(f(x, y)) → MINUS(y)
MINUS(f(x, y)) → MINUS(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(f(x1, x2)) = 4 + (4)x_1 + (4)x_2   
POL(h(x1)) = 4 + (4)x_1   
POL(MINUS(x1)) = (4)x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ PisEmptyProof

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

minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(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.