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
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

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

(2) Obligation:

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

(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:

ODD(s(x)) → NOT(odd(x))
ODD(s(x)) → ODD(x)
+¹(x, s(y)) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

+¹(s(x), y) → +¹(x, y)
+¹(x, s(y)) → +¹(x, y)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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.

+¹(x, s(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2)  =  x2
s(x1)  =  s(x1)
not(x1)  =  not
true  =  true
false  =  false
odd(x1)  =  odd
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

odd > not > false
odd > not > true
+₂ > s₁

Status:

trivial

The following usable rules [FROCOS05] were oriented:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

(9) Obligation:

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

+¹(s(x), y) → +¹(x, y)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(s(x), y) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2)  =  x1
s(x1)  =  s(x1)
not(x1)  =  not
true  =  true
false  =  false
odd(x1)  =  odd
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

odd > not > false
odd > not > true
+₂ > s₁

Status:

trivial

The following usable rules [FROCOS05] were oriented:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

(11) Obligation:

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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

ODD(s(x)) → ODD(x)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ODD(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)  =  x1
s(x1)  =  s(x1)
not(x1)  =  not
true  =  true
false  =  false
odd(x1)  =  odd
0  =  0
+(x1, x2)  =  +(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

odd > not > false
odd > not > true
+₂ > s₁

Status:

trivial

The following usable rules [FROCOS05] were oriented:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

(16) Obligation:

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

The set Q consists of the following terms:

not(true)
not(false)
odd(0)
odd(s(x0))
+(x0, 0)
+(x0, s(x1))
+(s(x0), x1)

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE