Term Rewriting System R:

   [x, y]
   f(x, x) -> f(i(x), g(g(x)))
   f(x, y) -> x
   f(x, i(x)) -> f(x, x)
   f(i(x), i(g(x))) -> a
   g(x) -> i(x)

Termination of R to be shown.


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

   f(x, y) -> x
   f(i(x), i(g(x))) -> a

where the Polynomial interpretation:
POL(g(x_1)) = x_1
POL(i(x_1)) = x_1
POL(a) = 0
POL(f(x_1, x_2)) = 1 + x_1 + x_2
was used. 



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


This program has no overlaps, so it is sufficient to show innermost termination. 


R contains the following Dependency Pairs: 


   F(x, i(x)) -> F(x, x)
   F(x, x) -> F(i(x), g(g(x)))
   F(x, x) -> G(g(x))
   F(x, x) -> G(x)

Furthermore, R contains one SCC.


SCC1:

F(x, x) -> F(i(x), g(g(x)))
F(x, i(x)) -> F(x, x)

On this Scc, a Rewriting SCC transformation can be performed.
 
As a result of transforming the rule 

   F(x, x) -> F(i(x), g(g(x)))
one new Dependency Pair
is created:


   F(x, x) -> F(i(x), i(g(x)))

The transformation is resulting in one subcycle:


SCC1.Rewr1:

F(x, x) -> F(i(x), i(g(x)))
F(x, i(x)) -> F(x, x)

On this Scc, a Rewriting SCC transformation can be performed.
 
As a result of transforming the rule 

   F(x, x) -> F(i(x), i(g(x)))
one new Dependency Pair
is created:


   F(x, x) -> F(i(x), i(i(x)))

The transformation is resulting in one subcycle:


SCC1.Rewr1.Rewr1:

F(x, x) -> F(i(x), i(i(x)))
F(x, i(x)) -> F(x, x)

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(i(x_1)) = x_1
POL(F(x_1, x_2)) = 1 + x_1 + x_2

No Dependency Pairs can be deleted.

The following rules of R can be deleted: 

   g(x) -> i(x)


 This transformation is resulting in one new subcycle:


SCC1.Rewr1.Rewr1.MRR1:

F(x, i(x)) -> F(x, x)
F(x, x) -> F(i(x), i(i(x)))

Found an infinite P-chain over R:
P = F(x, i(x)) -> F(x, x)
F(x, x) -> F(i(x), i(i(x)))
R = []
s = F(x'', i(x'')) evaluates to t = F(i(x''), i(i(x'')))
Thus, s starts an infinite reduction as s matches t.


Non-Termination of R could be shown.

Duration: 0.566 seconds.