Term Rewriting System R:
[x, u, v, z, y]
admit(x, nil) -> nil
admit(x, .(u, .(v, .(w, z)))) -> cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) -> y

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

ADMIT(x, .(u, .(v, .(w, z)))) -> COND(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
ADMIT(x, .(u, .(v, .(w, z)))) -> ADMIT(carry(x, u, v), z)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pair:

ADMIT(x, .(u, .(v, .(w, z)))) -> ADMIT(carry(x, u, v), z)


Rules:


admit(x, nil) -> nil
admit(x, .(u, .(v, .(w, z)))) -> cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) -> y


Strategy:

innermost




The following dependency pair can be strictly oriented:

ADMIT(x, .(u, .(v, .(w, z)))) -> ADMIT(carry(x, u, v), z)


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(carry(x1, x2, x3))=  0  
  POL(.(x1, x2))=  x1 + x2  
  POL(w)=  1  
  POL(ADMIT(x1, x2))=  1 + x2  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


admit(x, nil) -> nil
admit(x, .(u, .(v, .(w, z)))) -> cond(=(sum(x, u, v), w), .(u, .(v, .(w, admit(carry(x, u, v), z)))))
cond(true, y) -> y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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