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:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(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

neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))

The signature Sigma is {cond1, cond2}

(2) Obligation:

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

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(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:

COND1(true, x) → COND2(even(x), x)
COND1(true, x) → EVEN(x)
COND2(true, x) → COND1(neq(x, 0), div2(x))
COND2(true, x) → NEQ(x, 0)
COND2(true, x) → DIV2(x)
COND2(false, x) → COND1(neq(x, 0), p(x))
COND2(false, x) → NEQ(x, 0)
COND2(false, x) → P(x)
NEQ(s(x), s(y)) → NEQ(x, y)
EVEN(s(s(x))) → EVEN(x)
DIV2(s(s(x))) → DIV2(x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(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 3 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DIV2(s(s(x))) → DIV2(x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(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.

DIV2(s(s(x))) → DIV2(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DIV2(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(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:

EVEN(s(s(x))) → EVEN(x)

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

EVEN(s(s(x))) → EVEN(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
EVEN(x1)  =  x1
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

COND2(true, x) → COND1(neq(x, 0), div2(x))
COND1(true, x) → COND2(even(x), x)
COND2(false, x) → COND1(neq(x, 0), p(x))

The TRS R consists of the following rules:

cond1(true, x) → cond2(even(x), x)
cond2(true, x) → cond1(neq(x, 0), div2(x))
cond2(false, x) → cond1(neq(x, 0), p(x))
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
even(0) → true
even(s(0)) → false
even(s(s(x))) → even(x)
div2(0) → 0
div2(s(0)) → 0
div2(s(s(x))) → s(div2(x))
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(y))
even(0)
even(s(0))
even(s(s(x0)))
div2(0)
div2(s(0))
div2(s(s(x0)))
p(0)
p(s(x0))

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