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 oriented strictly and are deleted.


MINUS1(h1(x)) -> MINUS1(x)
The remaining pairs can at least be oriented weakly.

MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS1(x1) ) = max{0, x1 - 1}


POL( h1(x1) ) = x1 + 2


POL( f2(x1, x2) ) = x1 + x2 + 1



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(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 oriented strictly and are deleted.


MINUS1(f2(x, y)) -> MINUS1(x)
MINUS1(f2(x, y)) -> MINUS1(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS1(x1) ) = max{0, x1 - 1}


POL( f2(x1, x2) ) = x1 + x2 + 2



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.