Term Rewriting System R:
[x]
a(f, 0) -> a(s, 0)
a(d, 0) -> 0
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x

Innermost Termination of R to be shown. 



   R
     ↳Removing Redundant Rules



Removing the following rules from R which fullfill a polynomial ordering:

a(f, 0) -> a(s, 0)

where the Polynomial interpretation:
  POL(0)=  0  
  POL(d)=  0  
  POL(s)=  0  
  POL(a(x1, x2))=  x1 + x2  
  POL(f)=  1  
  POL(p)=  0  
was used.

Not all Rules of R can be deleted, so we still have to regard a part of R. 


   R
     ↳RRRPolo
       →TRS2
         ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

A(f, a(s, x)) -> A(d, a(f, a(p, a(s, x))))
A(f, a(s, x)) -> A(f, a(p, a(s, x)))
A(f, a(s, x)) -> A(p, a(s, x))
A(d, a(s, x)) -> A(s, a(s, a(d, a(p, a(s, x)))))
A(d, a(s, x)) -> A(s, a(d, a(p, a(s, x))))
A(d, a(s, x)) -> A(d, a(p, a(s, x)))
A(d, a(s, x)) -> A(p, a(s, x))

Furthermore, R contains two SCCs. 


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Usable Rules (Innermost)
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

A(d, a(s, x)) -> A(d, a(p, a(s, x)))


Rules:


a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(d, 0) -> 0


Strategy:

innermost




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


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 3
                 ↳A-Transformation
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

A(d, a(s, x)) -> A(d, a(p, a(s, x)))


Rule:


a(p, a(s, x)) -> x


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 4
                 ↳Modular Removal of Rules
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

D(s(x)) -> D(p(s(x)))


Rule:


p(s(x)) -> x


Strategy:

innermost




We have the following set of usable rules:

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

We have the following set D of usable symbols: {D, s, p}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

p(s(x)) -> x


The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 5
                 ↳Dependency Graph
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

D(s(x)) -> D(p(s(x)))


Rule:

none


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳Usable Rules (Innermost)


Dependency Pair:

A(f, a(s, x)) -> A(f, a(p, a(s, x)))


Rules:


a(f, a(s, x)) -> a(d, a(f, a(p, a(s, x))))
a(p, a(s, x)) -> x
a(d, a(s, x)) -> a(s, a(s, a(d, a(p, a(s, x)))))
a(d, 0) -> 0


Strategy:

innermost




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


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 6
                 ↳A-Transformation


Dependency Pair:

A(f, a(s, x)) -> A(f, a(p, a(s, x)))


Rule:


a(p, a(s, x)) -> x


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 7
                 ↳Modular Removal of Rules


Dependency Pair:

F(s(x)) -> F(p(s(x)))


Rule:


p(s(x)) -> x


Strategy:

innermost




We have the following set of usable rules:

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

We have the following set D of usable symbols: {s, F, p}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

p(s(x)) -> x


The result of this processor delivers one new DP problem.



   R
     ↳RRRPolo
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:

F(s(x)) -> F(p(s(x)))


Rule:

none


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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