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
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(x))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), 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:

MINUS1(h1(x)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), minus1(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:

MINUS1(h1(x)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), 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 strictly oriented and are deleted.


MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)
The remaining pairs can at least by weakly be oriented.

MINUS1(h1(x)) -> MINUS1(x)
Used ordering: Combined order from the following AFS and order.
MINUS1(x1)  =  MINUS1(x1)
h1(x1)  =  x1
f2(x1, x2)  =  f2(x1, x2)

Lexicographic Path Order [19].
Precedence:
[MINUS1, f2]


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:

MINUS1(h1(x)) -> MINUS1(x)

The TRS R consists of the following rules:

minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), 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 strictly oriented and are deleted.


MINUS1(h1(x)) -> MINUS1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS1(x1)  =  MINUS1(x1)
h1(x1)  =  h1(x1)

Lexicographic Path Order [19].
Precedence:
h1 > MINUS1


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:

minus1(minus1(x)) -> x
minus1(h1(x)) -> h1(minus1(x))
minus1(f2(x, y)) -> f2(minus1(y), 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.