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:

not1(x) -> xor2(x, true)
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
=2(x, y) -> xor2(x, xor2(y, true))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not1(x) -> xor2(x, true)
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
=2(x, y) -> xor2(x, xor2(y, true))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

not1(x) -> xor2(x, true)
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
=2(x, y) -> xor2(x, xor2(y, true))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ PisEmptyProof

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

not1(x) -> xor2(x, true)
implies2(x, y) -> xor2(and2(x, y), xor2(x, true))
or2(x, y) -> xor2(and2(x, y), xor2(x, y))
=2(x, y) -> xor2(x, xor2(y, true))

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.