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:

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 {cond2, cond1}

(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) 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:

DIV2(s(s(x))) → DIV2(x)

R is empty.
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) 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].

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))

(11) Obligation:

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

DIV2(s(s(x))) → DIV2(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:

• DIV2(s(s(x))) → DIV2(x)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) 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.

(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:

EVEN(s(s(x))) → EVEN(x)

R is empty.
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.

(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].

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))

(18) Obligation:

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

EVEN(s(s(x))) → EVEN(x)

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

(19) 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:

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

(20) TRUE

(21) 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.

(22) 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.

(23) 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:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
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))

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.

(24) 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].

cond1(true, x0)
cond2(true, x0)
cond2(false, x0)

(25) 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:

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
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))

The set Q consists of the following terms:

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.

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

POL(COND2(x₁, x₂)) = 3/2 + (5/4)x₂   
POL(true) = 4   
POL(COND1(x₁, x₂)) = (1/2)x₁ + (5/4)x₂   
POL(neq(x₁, x₂)) = x₁ + (1/4)x₂   
POL(0) = 1/4   
POL(div2(x₁)) = 1/2 + (1/2)x₁   
POL(even(x₁)) = 0   
POL(false) = 0   
POL(p(x₁)) = 1/4 + (1/4)x₁   
POL(s(x₁)) = 4 + (4)x₁   

The value of delta used in the strict ordering is 1/2.
The following usable rules [FROCOS05] were oriented:

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

(27) Obligation:

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

neq(0, 0) → false
neq(s(x), 0) → true
p(0) → 0
p(s(x)) → x
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))

The set Q consists of the following terms:

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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE