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,l1:S)
hd(cons(x:S,l:S)) -> x:S
ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S))
ifappend(l1:S,l2:S,nil) -> 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,l1:S)
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S))
 ifappend(l1:S,l2:S,nil) -> 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,l1:S)
 IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S)
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S))
 ifappend(l1:S,l2:S,nil) -> 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,l1:S)
 IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S)
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S))
 ifappend(l1:S,l2:S,nil) -> 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,l1:S)
 IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S)
->->-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S)
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S))
 ifappend(l1:S,l2:S,nil) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S

Problem 1: 

Subterm Processor:
-> Pairs:
 APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,l1:S)
 IFAPPEND(l1:S,l2:S,cons(x:S,l:S)) -> APPEND(l:S,l2:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S)
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S))
 ifappend(l1:S,l2:S,nil) -> l2:S
 is_empty(cons(x:S,l:S)) -> ffalse
 is_empty(nil) -> ttrue
 tl(cons(x:S,l:S)) -> l:S
->Projection:
 pi(APPEND) = 1
 pi(IFAPPEND) = 3

Problem 1: 

SCC Processor:
-> Pairs:
 APPEND(l1:S,l2:S) -> IFAPPEND(l1:S,l2:S,l1:S)
-> Rules:
 append(l1:S,l2:S) -> ifappend(l1:S,l2:S,l1:S)
 hd(cons(x:S,l:S)) -> x:S
 ifappend(l1:S,l2:S,cons(x:S,l:S)) -> cons(x:S,append(l:S,l2:S))
 ifappend(l1:S,l2:S,nil) -> 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.