Termination w.r.t. Q of the given QTRS could not be shown:

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ODD(s(s(x))) → ODD(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ODD(x1)  =  ODD(x1)
s(x1)  =  s(x1)
cond(x1, x2)  =  cond
true  =  true
odd(x1)  =  odd(x1)
p(x1)  =  p(x1)
0  =  0
false  =  false

Lexicographic path order with status [LPO].
Precedence:

s₁ > ODD₁
cond > odd₁ > true > ODD₁
cond > odd₁ > false > ODD₁
p₁ > 0 > true > ODD₁

Status:

ODD₁: [1]
odd₁: [1]
true: []
cond: []
false: []
p₁: [1]
s₁: [1]
0: []

The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

Q DP problem:
P is empty.
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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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.