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:

cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond(true, x) → cond(odd(x), p(p(p(x))))

The signature Sigma is {cond}

(2) Obligation:

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

cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

COND(true, x) → COND(odd(x), p(p(p(x))))
COND(true, x) → ODD(x)
COND(true, x) → P(p(p(x)))
COND(true, x) → P(p(x))
COND(true, x) → P(x)
ODD(s(s(x))) → ODD(x)

The TRS R consists of the following rules:

cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

ODD(s(s(x))) → ODD(x)

The TRS R consists of the following rules:

cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

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

(8) 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.

(9) Obligation:

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

ODD(s(s(x))) → ODD(x)

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

cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

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

(10) 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].

cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

(11) Obligation:

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

ODD(s(s(x))) → ODD(x)

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

(12) 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:

• ODD(s(s(x))) → ODD(x)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

COND(true, x) → COND(odd(x), p(p(p(x))))

The TRS R consists of the following rules:

cond(true, x) → cond(odd(x), p(p(p(x))))
odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

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

(15) 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.

(16) Obligation:

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

COND(true, x) → COND(odd(x), p(p(p(x))))

The TRS R consists of the following rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond(true, x0)
odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

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

(17) 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].

cond(true, x0)

(18) Obligation:

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

COND(true, x) → COND(odd(x), p(p(p(x))))

The TRS R consists of the following rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

COND(true, x) → COND(odd(x), p(p(p(x))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(COND(x₁, x₂)) = (4)x₁ + (3)x₂   
POL(true) = 2   
POL(odd(x₁)) = (1/2)x₁   
POL(p(x₁)) = (1/4)x₁   
POL(0) = 0   
POL(s(x₁)) = 4 + (4)x₁   
POL(false) = 0   

The value of delta used in the strict ordering is 8.
The following usable rules [FROCOS05] were oriented:

p(0) → 0
odd(s(s(x))) → odd(x)
p(s(x)) → x
odd(s(0)) → true
odd(0) → false

(20) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

odd(0) → false
odd(s(0)) → true
odd(s(s(x))) → odd(x)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

odd(0)
odd(s(0))
odd(s(s(x0)))
p(0)
p(s(x0))

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

(21) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(22) TRUE