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) 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:

NOT(x) → XOR(x, true)
OR(x, y) → XOR(and(x, y), xor(x, y))
OR(x, y) → AND(x, y)
OR(x, y) → XOR(x, y)
IMPLIES(x, y) → XOR(and(x, y), xor(x, true))
IMPLIES(x, y) → AND(x, y)
IMPLIES(x, y) → XOR(x, true)
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)

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.
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 8 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:

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.
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)  =  x1
xor(x1, x2)  =  xor(x1, x2)
not(x1)  =  not(x1)
true  =  true
or(x1, x2)  =  or(x1, x2)
and(x1, x2)  =  x1
implies(x1, x2)  =  implies(x1, x2)
false  =  false

Lexicographic path order with status [LPO].
Precedence:

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

Status:

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

The following usable rules [FROCOS05] were oriented:

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))

(6) Obligation:

Q DP problem:
P is empty.
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.
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