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(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(0, s(0)) → false
evenodd(s(x), s(0)) → evenodd(x, 0)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(evenodd(x₁, x₂)) = 1 + x₁ + x₂   
POL(false) = 0   
POL(not(x₁)) = x₁   
POL(s(x₁)) = x₁   
POL(true) = 0   

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

evenodd(0, s(0)) → false

(2) Obligation:

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

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(s(x), s(0)) → evenodd(x, 0)

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(s(x0), s(0))

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EVENODD(x, 0) → NOT(evenodd(x, s(0)))
EVENODD(x, 0) → EVENODD(x, s(0))
EVENODD(s(x), s(0)) → EVENODD(x, 0)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(s(x0), s(0))

We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EVENODD(s(x), s(0)) → EVENODD(x, 0)
EVENODD(x, 0) → EVENODD(x, s(0))

The TRS R consists of the following rules:

not(true) → false
not(false) → true
evenodd(x, 0) → not(evenodd(x, s(0)))
evenodd(s(x), s(0)) → evenodd(x, 0)

The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(s(x0), s(0))

We have to consider all minimal (P,Q,R)-chains.

(9) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EVENODD(s(x), s(0)) → EVENODD(x, 0)
EVENODD(x, 0) → EVENODD(x, s(0))

R is empty.
The set Q consists of the following terms:

not(true)
not(false)
evenodd(x0, 0)
evenodd(s(x0), s(0))

We have to consider all minimal (P,Q,R)-chains.

(11) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

not(true)
not(false)
evenodd(x0, 0)
evenodd(s(x0), s(0))

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EVENODD(s(x), s(0)) → EVENODD(x, 0)
EVENODD(x, 0) → EVENODD(x, s(0))

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• EVENODD(x, 0) → EVENODD(x, s(0))
  The graph contains the following edges 1 >= 1

• EVENODD(s(x), s(0)) → EVENODD(x, 0)
  The graph contains the following edges 1 > 1, 2 > 2

(14) TRUE