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
  ↳ 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)

Q is empty.

Using Dependency Pairs [1,15] 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)

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:

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)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


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: Polynomial interpretation [25,35]:

POL(band(x1, x2)) = 4 + x_2   
POL(AND(x1, x2, x3)) = (2)x_1 + (3)x_2 + (4)x_3   
POL(not(x1)) = 2 + (4)x_1   
The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none



↳ 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)

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.