Term Rewriting System R:
[x, y, z]
max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))
MAX(N(L(x), N(y, z))) -> MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pair:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))


Rules:


max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MAX(x1))=  1 + x1  
  POL(L(x1))=  x1  
  POL(N(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo


Dependency Pair:


Rules:


max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering


Dependency Pair:

MAX(N(L(x), N(y, z))) -> MAX(N(y, z))


Rules:


max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MAX(N(L(x), N(y, z))) -> MAX(N(y, z))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MAX(x1))=  1 + x1  
  POL(L(x1))=  0  
  POL(N(x1, x2))=  1 + x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph


Dependency Pair:


Rules:


max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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