Term Rewriting System R:
[y, x, u, v, z]
merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if(< (x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MERGE(.(x, y), .(u, v)) -> IF(< (x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
MERGE(.(x, y), .(u, v)) -> MERGE(y, .(u, v))
MERGE(.(x, y), .(u, v)) -> MERGE(.(x, y), v)
++'(.(x, y), z) -> ++'(y, z)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pairs:

MERGE(.(x, y), .(u, v)) -> MERGE(.(x, y), v)
MERGE(.(x, y), .(u, v)) -> MERGE(y, .(u, v))


Rules:


merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if(< (x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x





The following dependency pair can be strictly oriented:

MERGE(.(x, y), .(u, v)) -> MERGE(y, .(u, v))


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(.(x1, x2))=  1 + x2  
  POL(MERGE(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pair:

MERGE(.(x, y), .(u, v)) -> MERGE(.(x, y), v)


Rules:


merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if(< (x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x





The following dependency pair can be strictly oriented:

MERGE(.(x, y), .(u, v)) -> MERGE(.(x, y), v)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(.(x1, x2))=  1 + x2  
  POL(MERGE(x1, x2))=  x2  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if(< (x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering


Dependency Pair:

++'(.(x, y), z) -> ++'(y, z)


Rules:


merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if(< (x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x





The following dependency pair can be strictly oriented:

++'(.(x, y), z) -> ++'(y, z)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(++'(x1, x2))=  x1  
  POL(.(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph


Dependency Pair:


Rules:


merge(nil, y) -> y
merge(x, nil) -> x
merge(.(x, y), .(u, v)) -> if(< (x, u), .(x, merge(y, .(u, v))), .(u, merge(.(x, y), v)))
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))
if(true, x, y) -> x
if(false, x, y) -> x





Using the Dependency Graph resulted in no new DP problems.

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