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:

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

Q is empty.


↳ QTRS
  ↳ AAECC Innermost

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

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

The signature Sigma is {and}

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof

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

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

The set Q consists of the following terms:

and(not(not(x0)), x1, not(x2))


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

AND(not(not(x)), y, not(z)) → AND(y, band(x, z), x)

The TRS R consists of the following rules:

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

The set Q consists of the following terms:

and(not(not(x0)), x1, not(x2))

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

↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof

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

AND(not(not(x)), y, not(z)) → AND(y, band(x, z), x)

The TRS R consists of the following rules:

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

The set Q consists of the following terms:

and(not(not(x0)), x1, not(x2))

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.


AND(not(not(x)), y, not(z)) → AND(y, band(x, z), x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
AND(x1, x2, x3)  =  AND(x1, x2)
not(x1)  =  not
band(x1, x2)  =  band

Recursive Path Order [2].
Precedence:
not > AND2
not > band

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ AAECC Innermost
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

and(not(not(x)), y, not(z)) → and(y, band(x, z), x)

The set Q consists of the following terms:

and(not(not(x0)), x1, not(x2))

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