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:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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(xor(x, y), z) → AND(x, z)
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
IMPL(x, y) → AND(x, y)
IMPL(x, y) → XOR(and(x, y), xor(x, T))
EQUIV(x, y) → XOR(y, T)
IMPL(x, y) → XOR(x, T)
OR(x, y) → XOR(x, y)
AND(xor(x, y), z) → AND(y, z)
NEG(x) → XOR(x, T)
EQUIV(x, y) → XOR(x, xor(y, T))
OR(x, y) → AND(x, y)
OR(x, y) → XOR(and(x, y), xor(x, y))

The TRS R consists of the following rules:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

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:

AND(xor(x, y), z) → AND(x, z)
AND(xor(x, y), z) → XOR(and(x, z), and(y, z))
IMPL(x, y) → AND(x, y)
IMPL(x, y) → XOR(and(x, y), xor(x, T))
EQUIV(x, y) → XOR(y, T)
IMPL(x, y) → XOR(x, T)
OR(x, y) → XOR(x, y)
AND(xor(x, y), z) → AND(y, z)
NEG(x) → XOR(x, T)
EQUIV(x, y) → XOR(x, xor(y, T))
OR(x, y) → AND(x, y)
OR(x, y) → XOR(and(x, y), xor(x, y))

The TRS R consists of the following rules:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 10 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

AND(xor(x, y), z) → AND(x, z)
AND(xor(x, y), z) → AND(y, z)

The TRS R consists of the following rules:

xor(x, F) → x
xor(x, neg(x)) → F
and(x, T) → x
and(x, F) → F
and(x, x) → x
and(xor(x, y), z) → xor(and(x, z), and(y, z))
xor(x, x) → F
impl(x, y) → xor(and(x, y), xor(x, T))
or(x, y) → xor(and(x, y), xor(x, y))
equiv(x, y) → xor(x, xor(y, T))
neg(x) → xor(x, T)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QDPSizeChangeProof

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

AND(xor(x, y), z) → AND(x, z)
AND(xor(x, y), z) → AND(y, z)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: