Term Rewriting System R:
[x, l, l1, l2]
isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, l1)
ifappend(l1, l2, nil) -> l2
ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APPEND(l1, l2) -> IFAPPEND(l1, l2, l1)
IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering


Dependency Pairs:

IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2)
APPEND(l1, l2) -> IFAPPEND(l1, l2, l1)


Rules:


isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, l1)
ifappend(l1, l2, nil) -> l2
ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFAPPEND(l1, l2, cons(x, l)) -> APPEND(l, l2)


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(IFAPPEND(x1, x2, x3))=  x3  
  POL(cons(x1, x2))=  1 + x2  
  POL(APPEND(x1, x2))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

APPEND(l1, l2) -> IFAPPEND(l1, l2, l1)


Rules:


isempty(nil) -> true
isempty(cons(x, l)) -> false
hd(cons(x, l)) -> x
tl(cons(x, l)) -> l
append(l1, l2) -> ifappend(l1, l2, l1)
ifappend(l1, l2, nil) -> l2
ifappend(l1, l2, cons(x, l)) -> cons(x, append(l, l2))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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