Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

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.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 10 less nodes.

(4) Obligation:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2)  =  AND(x1, x2)
xor(x1, x2)  =  xor(x1, x2)
F  =  F
neg(x1)  =  neg(x1)
and(x1, x2)  =  x1
T  =  T
impl(x1, x2)  =  impl(x1, x2)
or(x1, x2)  =  or(x1, x2)
equiv(x1, x2)  =  equiv(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

AND₂ > F
neg₁ > xor₂ > F
neg₁ > T > F
impl₂ > xor₂ > F
impl₂ > T > F
or₂ > xor₂ > F
equiv₂ > xor₂ > F
equiv₂ > T > F

Status:

AND₂: [1,2]
xor₂: [1,2]
F: []
neg₁: [1]
T: []
impl₂: [1,2]
or₂: [1,2]
equiv₂: [1,2]

The following usable rules [FROCOS05] were oriented:

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)

(6) Obligation:

Q DP problem:
P is empty.
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.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE