Termination w.r.t. Q proof of TRS/SK902.30.trs

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:

not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))

The set Q consists of the following terms:

not(x0)
implies(x0, x1)
or(x0, x1)
=(x0, x1)

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:

not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))

The set Q consists of the following terms:

not(x0)
implies(x0, x1)
or(x0, x1)
=(x0, x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ PisEmptyProof

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

not(x) → xor(x, true)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))

The set Q consists of the following terms:

not(x0)
implies(x0, x1)
or(x0, x1)
=(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.