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].
Precedence:

not₁ > xor₂ > false
not₁ > true > false
or₂ > and₂ > xor₂ > false
implies₂ > true > false
implies₂ > and₂ > xor₂ > false

Status:

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

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