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:

and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))

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(not1(x), not1(y)) -> IMPLIES2(y, and2(x, y))
IMPLIES2(false, y) -> NOT1(false)
IMPLIES2(not1(x), not1(y)) -> AND2(x, y)
IMPLIES2(x, false) -> NOT1(x)

The TRS R consists of the following rules:

and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

IMPLIES2(not1(x), not1(y)) -> IMPLIES2(y, and2(x, y))
IMPLIES2(false, y) -> NOT1(false)
IMPLIES2(not1(x), not1(y)) -> AND2(x, y)
IMPLIES2(x, false) -> NOT1(x)

The TRS R consists of the following rules:

and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

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

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

IMPLIES2(not1(x), not1(y)) -> IMPLIES2(y, and2(x, y))

The TRS R consists of the following rules:

and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))

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.


IMPLIES2(not1(x), not1(y)) -> IMPLIES2(y, and2(x, y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( IMPLIES2(x1, x2) ) = max{0, x1 + x2 - 3}


POL( not1(x1) ) = x1 + 2


POL( and2(x1, x2) ) = x1


POL( false ) = 0



The following usable rules [14] were oriented:

and2(x, not1(false)) -> x
and2(x, false) -> false



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

and2(x, false) -> false
and2(x, not1(false)) -> x
not1(not1(x)) -> x
implies2(false, y) -> not1(false)
implies2(x, false) -> not1(x)
implies2(not1(x), not1(y)) -> implies2(y, and2(x, y))

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.