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:

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, z))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Quasi-Precedence:

[not₁, true, implies₂] > and₂ > [xor₂, false]
or₂ > and₂ > [xor₂, false]

Status:

xor₂: multiset
not₁: [1]
or₂: multiset
true: multiset
false: multiset
and₂: [2,1]
implies₂: [2,1]

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

not(x) → xor(x, true)
or(x, y) → xor(and(x, y), xor(x, y))
implies(x, y) → xor(and(x, y), xor(x, true))
and(x, true) → x
and(x, false) → false
and(x, x) → x
xor(x, false) → x
xor(x, x) → false
and(xor(x, y), z) → xor(and(x, z), and(y, z))

(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