Term Rewriting System R:

   [x, y, z]
   f(x, 0, 0) -> s(x)
   f(0, y, 0) -> s(y)
   f(0, 0, z) -> s(z)
   f(s(0), y, z) -> f(0, s(y), s(z))
   f(s(x), s(y), 0) -> f(x, y, s(0))
   f(s(x), 0, s(z)) -> f(x, s(0), z)
   f(0, s(0), s(0)) -> s(s(0))
   f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
   f(0, s(s(y)), s(0)) -> f(0, y, s(0))
   f(0, s(0), s(s(z))) -> f(0, s(0), z)
   f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Termination of R to be shown.


R contains the following Dependency Pairs: 


   F(0, s(s(y)), s(0)) -> F(0, y, s(0))
   F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))
   F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
   F(s(x), s(y), 0) -> F(x, y, s(0))
   F(s(x), 0, s(z)) -> F(x, s(0), z)
   F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
   F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
   F(0, s(0), s(s(z))) -> F(0, s(0), z)
   F(s(0), y, z) -> F(0, s(y), s(z))

Furthermore, R contains four SCCs.


SCC1:

F(0, s(s(y)), s(0)) -> F(0, y, s(0))

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(s(x_1)) = 1 + x_1
POL(F(x_1, x_2, x_3)) = 1 + x_1 + x_2 + x_3
POL(0) = 1

The following Dependency Pairs can be deleted:

   F(0, s(s(y)), s(0)) -> F(0, y, s(0))



 This transformation is resulting in no new subcycles.


SCC2:

F(0, s(0), s(s(z))) -> F(0, s(0), z)

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(s(x_1)) = 1 + x_1
POL(F(x_1, x_2, x_3)) = 1 + x_1 + x_2 + x_3
POL(0) = 1

The following Dependency Pairs can be deleted:

   F(0, s(0), s(s(z))) -> F(0, s(0), z)



 This transformation is resulting in no new subcycles.


SCC3:

F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))

By using a polynomial ordering, at least one Dependency Pair of this SCC can be strictly oriented.

No rules need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(F(x_1, x_2, x_3)) = x_2
POL(f(x_1, x_2, x_3)) = 0
POL(0) = 0

 resulting in one subcycle.


SCC3.Polo1:

F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))

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(s(x_1)) = 1 + x_1
POL(F(x_1, x_2, x_3)) = 1 + x_1 + x_2 + x_3
POL(0) = 1

The following Dependency Pairs can be deleted:

   F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))



 This transformation is resulting in no new subcycles.


SCC4:

F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))

By using a polynomial ordering, at least one Dependency Pair of this SCC can be strictly oriented.

No rules need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
POL(s(x_1)) = 1 + x_1
POL(F(x_1, x_2, x_3)) = x_1
POL(f(x_1, x_2, x_3)) = 0
POL(0) = 0

 resulting in one subcycle.


SCC4.Polo1:

F(s(x), s(y), s(z)) -> F(s(x), s(y), z)

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(s(x_1)) = 1 + x_1
POL(F(x_1, x_2, x_3)) = 1 + x_1 + x_2 + x_3

The following Dependency Pairs can be deleted:

   F(s(x), s(y), s(z)) -> F(s(x), s(y), z)



 This transformation is resulting in no new subcycles.




Termination of R successfully shown.


Duration: 1.99 seconds.