Term Rewriting System R:
[x, y, z]
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(x, s(y)) -> +'(x, y)
+'(s(x), y) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
F(g(f(x))) -> F(h(s(0), x))
F(g(h(x, y))) -> F(h(s(x), y))
F(h(x, h(y, z))) -> F(h(+(x, y), z))
F(h(x, h(y, z))) -> +'(x, y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))


Strategy:

innermost




As we are in the innermost case, we can delete all 3 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 3
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules


Dependency Pairs:

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


Rules:


+(x, 0) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)


Strategy:

innermost




We number the DPs as follows:
  1. +'(x, +(y, z)) -> +'(x, y)
  2. +'(x, +(y, z)) -> +'(+(x, y), z)
  3. +'(s(x), y) -> +'(x, y)
  4. +'(x, s(y)) -> +'(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2
{2} , {2}
2>2
{3} , {3}
1>1
2=2
{4} , {4}
1=1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2
{3} , {3}
1>1
2=2
{4} , {4}
1=1
2>2
{3} , {2}
2>2
{3} , {4}
1>1
2>2
{4} , {3}
1>1
2>2
{2} , {3}
2>2
{1} , {3}
1>1
2>2
{3} , {1}
1>1
2>2
{2} , {1}
2>2
{1} , {1}
1>1
2>2
{3} , {3}
2>2
{4} , {4}
1>1
2>2
{1} , {4}
1>1
2>2
{4} , {3}
2>2
{3} , {3}
1>1
2>2
{1} , {3}
2>2
{2} , {4}
2>2
{4} , {4}
2>2
{1} , {4}
2>2
{1} , {1}
2>2
{3} , {1}
2>2
{3} , {4}
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)


Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))


Rules:


+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))


Strategy:

innermost




As we are in the innermost case, we can delete all 3 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳Modular Removal of Rules


Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))


Rules:


+(x, 0) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)


Strategy:

innermost




We have the following set of usable rules:

+(x, 0) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(0)=  1  
  POL(h(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  
  POL(F(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {h, s, F, +}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

+(x, 0) -> x
+(0, y) -> y


The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 4
             ↳MRR
             ...
               →DP Problem 5
                 ↳Modular Removal of Rules


Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))


Rules:


+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)


Strategy:

innermost




We have the following set of usable rules:

+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(h(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  x1  
  POL(F(x1))=  1 + x1  
  POL(+(x1, x2))=  x1 + x2  

We have the following set D of usable symbols: {h, s, F, +}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

F(h(x, h(y, z))) -> F(h(+(x, y), z))

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.


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