Term Rewriting System R:
[x, y, z]
double(0) -> 0
double(s(x)) -> s(s(double(x)))
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
half(double(x)) -> x
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
if(0, y, z) -> y
if(s(x), y, z) -> z

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

DOUBLE(s(x)) -> DOUBLE(x)
HALF(s(s(x))) -> HALF(x)
-'(s(x), s(y)) -> -'(x, y)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)


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





The following dependency pair can be strictly oriented:

DOUBLE(s(x)) -> DOUBLE(x)


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
DOUBLE(x1) -> DOUBLE(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS


Dependency Pair:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳AFS


Dependency Pair:

HALF(s(s(x))) -> HALF(x)


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





The following dependency pair can be strictly oriented:

HALF(s(s(x))) -> HALF(x)


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
HALF(x1) -> HALF(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳AFS


Dependency Pair:


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳Argument Filtering and Ordering


Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)


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





The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(x, y)


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
-'(x1, x2) -> -'(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
       →DP Problem 3
         ↳AFS
           →DP Problem 6
             ↳Dependency Graph


Dependency Pair:


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





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes