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(x1, x2)
s(x1)  =  s(x1)
double(x1)  =  double(x1)
0  =  0
half(x1)  =  half(x1)
-(x1, x2)  =  -(x1, x2)
if(x1, x2, x3)  =  if(x2, x3)

Recursive Path Order [RPO].
Precedence:
double1 > [s1, half1] > -^12
double1 > [s1, half1] > 0


The following usable rules [FROCOS05] were oriented:

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

(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)
double(x1)  =  double(x1)
0  =  0
half(x1)  =  half(x1)
-(x1, x2)  =  -(x1)
if(x1, x2, x3)  =  if(x2, x3)

Recursive Path Order [RPO].
Precedence:
[double1, 0, half1] > s1 > -1


The following usable rules [FROCOS05] were oriented:

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

(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)
double(x1)  =  double(x1)
0  =  0
half(x1)  =  half(x1)
-(x1, x2)  =  -(x1, x2)
if(x1, x2, x3)  =  if(x1, x2, x3)

Recursive Path Order [RPO].
Precedence:
[double1, 0] > [s1, -2] > half1
if3 > half1


The following usable rules [FROCOS05] were oriented:

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

(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