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 [1,13] 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(x)
MINUS(f(x, y)) → MINUS(y)

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(x)
MINUS(f(x, y)) → MINUS(y)

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 [13].


The following pairs can be oriented strictly and are deleted.


MINUS(f(x, y)) → MINUS(x)
MINUS(f(x, y)) → MINUS(y)
The remaining pairs can at least be oriented weakly.

MINUS(h(x)) → MINUS(x)
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  MINUS(x1)
h(x1)  =  x1
f(x1, x2)  =  f(x1, x2)

Recursive path order with status [2].
Precedence:
f2 > MINUS1

Status:
f2: multiset
MINUS1: multiset

The following usable rules [14] were oriented: none



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

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

MINUS(h(x)) → 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 [13].


The following pairs can be oriented strictly and are deleted.


MINUS(h(x)) → MINUS(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  MINUS(x1)
h(x1)  =  h(x1)

Recursive path order with status [2].
Precedence:
h1 > MINUS1

Status:
h1: multiset
MINUS1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ 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.