Term Rewriting System R:
[x]
f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(a, x) -> F(b, f(c, x))
F(a, x) -> F(c, x)
F(a, f(b, x)) -> F(b, f(a, x))
F(a, f(b, x)) -> F(a, x)
F(d, f(c, x)) -> F(d, f(a, x))
F(d, f(c, x)) -> F(a, x)
F(a, f(c, x)) -> F(c, f(a, x))
F(a, f(c, x)) -> F(a, x)

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

F(a, f(c, x)) -> F(a, x)
F(a, f(b, x)) -> F(a, x)


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))





The original DP problem is in applicative form. Its DPs and usable rules are the following.

F(a, f(c, x)) -> F(a, x)
F(a, f(b, x)) -> F(a, x)

none


It is proper and hence, it can be A-transformed which results in the DP problem

A(c(x)) -> A(x)
A(b(x)) -> A(x)

none


We number the DPs as follows:
  1. A(c(x)) -> A(x)
  2. A(b(x)) -> A(x)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1

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

DP: empty set
Oriented Rules: none

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

with Argument Filtering System:
c(x1) -> c(x1)
b(x1) -> b(x1)

We obtain no new DP problems. 


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


Dependency Pair:

F(d, f(c, x)) -> F(d, f(a, x))


Rules:


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(d, f(c, x)) -> f(d, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))





The original DP problem is in applicative form. Its DPs and usable rules are the following.

F(d, f(c, x)) -> F(d, f(a, x))


f(a, x) -> f(b, f(c, x))
f(a, f(b, x)) -> f(b, f(a, x))
f(a, f(c, x)) -> f(c, f(a, x))


It is proper and hence, it can be A-transformed which results in the DP problem

D(c(x)) -> D(a(x))


a(x) -> b(c(x))
a(b(x)) -> b(a(x))
a(c(x)) -> c(a(x))


To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(c(x1))=  x1  
  POL(b(x1))=  x1  
  POL(D(x1))=  x1  
  POL(a(x1))=  x1  

We have the following set D of usable symbols: {c, b, D, a}
No Dependency Pairs can be deleted.
1 non usable rules have been deleted.

The result of this processor delivers one new DP problem.
Note that we keep the A-transformed DP problem as result of this processor.



   R
     ↳DPs
       →DP Problem 1
         ↳SCP
       →DP Problem 2
         ↳MRR
           →DP Problem 3
             ↳Negative Polynomial Order


Dependency Pair:

D(c(x)) -> D(a(x))


Rules:


a(x) -> b(c(x))
a(b(x)) -> b(a(x))
a(c(x)) -> c(a(x))





The following Dependency Pair can be strictly oriented using the given order.

D(c(x)) -> D(a(x))


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

a(x) -> b(c(x))
a(b(x)) -> b(a(x))
a(c(x)) -> c(a(x))


Used ordering:
Polynomial Order with Interpretation:

POL( D(x1) ) = x1

POL( c(x1) ) = x1 + 1

POL( a(x1) ) = x1

POL( b(x1) ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳SCP
       →DP Problem 2
         ↳MRR
           →DP Problem 3
             ↳Neg POLO
             ...
               →DP Problem 4
                 ↳Dependency Graph


Dependency Pair:


Rules:


a(x) -> b(c(x))
a(b(x)) -> b(a(x))
a(c(x)) -> c(a(x))





Using the Dependency Graph resulted in no new DP problems.

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