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)

Innermost Termination of R to be shown. 



   R
     ↳Removing Redundant Rules for Innermost Termination



Removing the following rules from R which left hand sides contain non normal subterms

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


   R
     ↳RRRI
       →TRS2
         ↳Removing Redundant Rules



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

f(x, y) -> x

where the Polynomial interpretation:
  POL(i(x1))=  x1  
  POL(g(x1))=  x1  
  POL(f(x1, x2))=  1 + x1 + x2  
was used.

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


   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳Dependency Pair Analysis



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. 


   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 1
                 ↳Non-Overlappingness Check


Dependency Pairs:

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


Rules:


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





R does not overlap into P. Moreover, R is locally confluent (all critical pairs are trivially joinable).Hence we can switch to innermost.
The transformation is resulting in one subcycle: 


   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 2
                 ↳Dependency Graph


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 1 DP problems.


   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 3
                 ↳Usable Rules (Innermost)


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




As we are in the innermost case, we can delete all 2 non-usable-rules.


   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 4
                 ↳Rewriting Transformation


Dependency Pairs:

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


Rule:


g(x) -> i(x)


Strategy:

innermost




On this DP problem, 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 new DP problem: 



   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 5
                 ↳Rewriting Transformation


Dependency Pairs:

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


Rule:


g(x) -> i(x)


Strategy:

innermost




On this DP problem, 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 new DP problem: 



   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 6
                 ↳Usable Rules (Innermost)


Dependency Pairs:

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


Rule:


g(x) -> i(x)


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳RRRI
       →TRS2
         ↳RRRPolo
           →TRS3
             ↳DPs
             ...
               →DP Problem 7
                 ↳Non Termination


Dependency Pairs:

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


Rule:

none


Strategy:

innermost




Found an infinite P-chain over R:
P =

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

R = none

s = F(x'', i(x''))
evaluates to t =F(i(x''), i(i(x'')))

Thus, s starts an infinite chain as s matches t.

Innermost Non-Termination of R could be shown.
Duration:
0:00 minutes