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)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Lexicographic path order with status [LPO].
Precedence:

not₁ > true > xor₂
implies₂ > true > xor₂
implies₂ > and₂ > xor₂
or₂ > and₂ > xor₂
=₂ > true > xor₂

Status:

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

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)
implies(x, y) → xor(and(x, y), xor(x, true))
or(x, y) → xor(and(x, y), xor(x, y))
=(x, y) → xor(x, xor(y, true))

(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