Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE

(0) Obligation:

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

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)
HALF(s(s(x))) → HALF(x)
-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs.

(4) Complex Obligation (AND)

(5) Obligation:

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

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


-1(s(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)  =  -1(x2)
s(x1)  =  s(x1)

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

Status:
-^11: [1]
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

HALF(s(s(x))) → HALF(x)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


HALF(s(s(x))) → HALF(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
HALF(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

(12) Obligation:

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

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DOUBLE(s(x)) → DOUBLE(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DOUBLE(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

(17) Obligation:

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

double(0) → 0
double(s(x)) → s(s(double(x)))
half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
if(0, y, z) → y
if(s(x), y, z) → z
half(double(x)) → x

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE