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

Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC.


   R
DPs
       →DP Problem 1
Modular Removal of Rules


Dependency Pairs:

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


Rule:


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





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

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


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


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

A(a(b(a(a(b(a(x))))))) -> A(a(b(x)))
A(a(b(a(a(b(a(x))))))) -> A(a(a(b(x))))
A(a(b(a(a(b(a(x))))))) -> A(a(b(a(a(a(b(x)))))))


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


To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(b(x1))=  x1  
  POL(a(x1))=  x1  
  POL(A(x1))=  1 + x1  

We have the following set D of usable symbols: {b, a, A}
No Dependency Pairs can be deleted.
No Rules can be 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
MRR
           →DP Problem 2
Narrowing Transformation


Dependency Pairs:

A(a(b(a(a(b(a(x))))))) -> A(a(b(x)))
A(a(b(a(a(b(a(x))))))) -> A(a(a(b(x))))
A(a(b(a(a(b(a(x))))))) -> A(a(b(a(a(a(b(x)))))))


Rule:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

A(a(b(a(a(b(a(x))))))) -> A(a(a(b(x))))
one new Dependency Pair is created:

A(a(b(a(a(b(a(a(a(b(a(x''))))))))))) -> A(a(b(a(a(b(a(a(a(b(x''))))))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
MRR
           →DP Problem 2
Nar
             ...
               →DP Problem 3
Narrowing Transformation


Dependency Pairs:

A(a(b(a(a(b(a(a(a(b(a(x''))))))))))) -> A(a(b(a(a(b(a(a(a(b(x''))))))))))
A(a(b(a(a(b(a(x))))))) -> A(a(b(a(a(a(b(x)))))))
A(a(b(a(a(b(a(x))))))) -> A(a(b(x)))


Rule:


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





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

A(a(b(a(a(b(a(x))))))) -> A(a(b(a(a(a(b(x)))))))
one new Dependency Pair is created:

A(a(b(a(a(b(a(a(a(b(a(x''))))))))))) -> A(a(b(a(a(b(a(a(b(a(a(a(b(x'')))))))))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
MRR
           →DP Problem 2
Nar
             ...
               →DP Problem 4
RFC Match Bounds


Dependency Pairs:

A(a(b(a(a(b(a(a(a(b(a(x''))))))))))) -> A(a(b(a(a(b(a(a(b(a(a(a(b(x'')))))))))))))
A(a(b(a(a(b(a(x))))))) -> A(a(b(x)))
A(a(b(a(a(b(a(a(a(b(a(x''))))))))))) -> A(a(b(a(a(b(a(a(a(b(x''))))))))))


Rule:


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





Using RFC Match Bounds, the DP problem could be solved. The Match Bound was 2.
The certificate found is represented by the following graph.

The certificate consists of the following enumerated nodes:

202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288

Node 202 is start node and node 203 is final node.

Those nodes are connect through the following edges:



Termination of R successfully shown.
Duration:
0:14 minutes