YES

Problem 1: 

(VAR v_NonEmpty:S l:S l1:S l2:S x:S)
(RULES
append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S))
hd(cons(x:S,l:S)) -> x:S
ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S))
ifappend(l1:S,l2:S,ttrue) -> l2:S
is_empty(cons(x:S,l:S)) -> ffalse
is_empty(nil) -> ttrue
tl(cons(x:S,l:S)) -> l:S
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S))
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S))
 ifappend(l1:S,l2:S,ttrue) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S))
 APPEND(l1:S,l2:S) -> IS_EMPTY(l1:S)
 IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S)
 IFAPPEND(l1:S,l2:S,ffalse) -> HD(l1:S)
 IFAPPEND(l1:S,l2:S,ffalse) -> TL(l1:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S))
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S))
 ifappend(l1:S,l2:S,ttrue) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S

Problem 1: 

SCC Processor:
-> Pairs:
 APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S))
 APPEND(l1:S,l2:S) -> IS_EMPTY(l1:S)
 IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S)
 IFAPPEND(l1:S,l2:S,ffalse) -> HD(l1:S)
 IFAPPEND(l1:S,l2:S,ffalse) -> TL(l1:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S))
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S))
 ifappend(l1:S,l2:S,ttrue) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S))
 IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S)
->->-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S))
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S))
 ifappend(l1:S,l2:S,ttrue) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,is_empty(l1:S))
 IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S))
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S))
 ifappend(l1:S,l2:S,ttrue) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S
-> Usable rules:
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[append](X1,X2) = 0
[hd](X) = 0
[ifappend](X1,X2,X3) = 0
[is_empty](X) = X
[tl](X) = 1/2.X
[cons](X1,X2) = 2.X2 + 2
[fSNonEmpty] = 0
[false] = 1/2
[nil] = 1/2
[true] = 1/2
[APPEND](X1,X2) = 2.X1 + 1
[HD](X) = 0
[IFAPPEND](X1,X2,X3) = X1 + X3 + 1/2
[IS_EMPTY](X) = 0
[TL](X) = 0

Problem 1: 

SCC Processor:
-> Pairs:
 IFAPPEND(l1:S,l2:S,ffalse) -> APPEND(tl(l1:S),l2:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,is_empty(l1:S))
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,ffalse) -> cons(hd(l1:S),append(tl(l1:S),l2:S))
 ifappend(l1:S,l2:S,ttrue) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.