Term Rewriting System R:
[x, y]
f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(c(s(x), y)) -> F(c(x, s(y)))
F(c(s(x), s(y))) -> G(c(x, y))
G(c(x, s(y))) -> G(c(s(x), y))
G(c(s(x), s(y))) -> F(c(x, y))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

G(c(s(x), s(y))) -> F(c(x, y))
G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), s(y))) -> G(c(x, y))
F(c(s(x), y)) -> F(c(x, s(y)))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

G(c(s(x), s(y))) -> F(c(x, y))
F(c(s(x), s(y))) -> G(c(x, y))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1, x2))=  x1 + x2  
  POL(G(x1))=  x1  
  POL(s(x1))=  1 + x1  
  POL(F(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳Dependency Graph


Dependency Pairs:

G(c(x, s(y))) -> G(c(s(x), y))
F(c(s(x), y)) -> F(c(x, s(y)))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 2 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 3
                 ↳Polynomial Ordering


Dependency Pair:

G(c(x, s(y))) -> G(c(s(x), y))


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

G(c(x, s(y))) -> G(c(s(x), y))


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1, x2))=  1 + x2  
  POL(G(x1))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 2
             ↳DGraph
             ...
               →DP Problem 4
                 ↳Polynomial Ordering


Dependency Pair:

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


Rules:


f(c(s(x), y)) -> f(c(x, s(y)))
f(c(s(x), s(y))) -> g(c(x, y))
g(c(x, s(y))) -> g(c(s(x), y))
g(c(s(x), s(y))) -> f(c(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(c(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(F(x1))=  1 + x1  

resulting in one new DP problem.


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