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) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive Path Order [RPO].
Precedence:

neg₁ > xor₂ > F
neg₁ > T
impl₂ > and₂ > xor₂ > F
impl₂ > T
or₂ > and₂ > xor₂ > F
equiv₂ > xor₂ > F
equiv₂ > T

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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)

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE