Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 QTRSRRRProof (⇔)
↳4 QTRS
  ↳5 RisEmptyProof (⇔)
    ↳6 TRUE
  ↳7 RisEmptyProof (⇔)
    ↳8 TRUE

(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:
Polynomial interpretation [POLO]:

POL(=(x1, x2)) = 2 + x1 + x2   
POL(and(x1, x2)) = 1 + x1 + x2   
POL(implies(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(not(x1)) = 2 + 2·x1   
POL(or(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(true) = 1   
POL(xor(x1, x2)) = x1 + x2   
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))
=(x, y) → xor(x, xor(y, true))


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(and(x1, x2)) = x1 + x2   
POL(implies(x1, x2)) = 1 + 2·x1 + 2·x2   
POL(true) = 0   
POL(xor(x1, x2)) = x1 + x2   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(4) Obligation:

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

(5) RisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE

(7) RisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE