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:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(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:

IMPLIES(x, or(y, z)) → IMPLIES(x, z)
IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

The TRS R consists of the following rules:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(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:

IMPLIES(x, or(y, z)) → IMPLIES(x, z)
IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

The TRS R consists of the following rules:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(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 oriented strictly and are deleted.


IMPLIES(x, or(y, z)) → IMPLIES(x, z)
The remaining pairs can at least be oriented weakly.

IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))
Used ordering: Combined order from the following AFS and order.
IMPLIES(x1, x2)  =  IMPLIES(x2)
or(x1, x2)  =  or(x2)
not(x1)  =  x1

Recursive path order with status [2].
Precedence:
trivial

Status:
or1: multiset
IMPLIES1: 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:

IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))

The TRS R consists of the following rules:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(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 oriented strictly and are deleted.


IMPLIES(not(x), or(y, z)) → IMPLIES(y, or(x, z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IMPLIES(x1, x2)  =  IMPLIES(x1, x2)
not(x1)  =  not(x1)
or(x1, x2)  =  or(x1)

Recursive path order with status [2].
Precedence:
not1 > IMPLIES2 > or1

Status:
IMPLIES2: multiset
or1: [1]
not1: 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:

implies(not(x), y) → or(x, y)
implies(not(x), or(y, z)) → implies(y, or(x, z))
implies(x, or(y, z)) → or(y, implies(x, z))

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.