Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

implies2(not1(x), y) -> or2(x, y)
implies2(not1(x), or2(y, z)) -> implies2(y, or2(x, z))
implies2(x, or2(y, z)) -> or2(y, implies2(x, z))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

implies2(not1(x), y) -> or2(x, y)
implies2(not1(x), or2(y, z)) -> implies2(y, or2(x, z))
implies2(x, or2(y, z)) -> or2(y, implies2(x, z))

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:

IMPLIES2(x, or2(y, z)) -> IMPLIES2(x, z)
IMPLIES2(not1(x), or2(y, z)) -> IMPLIES2(y, or2(x, z))

The TRS R consists of the following rules:

implies2(not1(x), y) -> or2(x, y)
implies2(not1(x), or2(y, z)) -> implies2(y, or2(x, z))
implies2(x, or2(y, z)) -> or2(y, implies2(x, z))

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:

IMPLIES2(x, or2(y, z)) -> IMPLIES2(x, z)
IMPLIES2(not1(x), or2(y, z)) -> IMPLIES2(y, or2(x, z))

The TRS R consists of the following rules:

implies2(not1(x), y) -> or2(x, y)
implies2(not1(x), or2(y, z)) -> implies2(y, or2(x, z))
implies2(x, or2(y, z)) -> or2(y, implies2(x, z))

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.


IMPLIES2(x, or2(y, z)) -> IMPLIES2(x, z)
The remaining pairs can at least by weakly be oriented.

IMPLIES2(not1(x), or2(y, z)) -> IMPLIES2(y, or2(x, z))
Used ordering: Combined order from the following AFS and order.
IMPLIES2(x1, x2)  =  IMPLIES1(x2)
or2(x1, x2)  =  or1(x2)
not1(x1)  =  x1

Lexicographic Path Order [19].
Precedence:
IMPLIES1 > or1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP

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

IMPLIES2(not1(x), or2(y, z)) -> IMPLIES2(y, or2(x, z))

The TRS R consists of the following rules:

implies2(not1(x), y) -> or2(x, y)
implies2(not1(x), or2(y, z)) -> implies2(y, or2(x, z))
implies2(x, or2(y, z)) -> or2(y, implies2(x, z))

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