YES

Problem 1: 

(VAR v_NonEmpty:S l:S x:S)
(RULES
conv(x:S) -> conviter(x:S,cons(0,nil))
conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x:S))) -> s(half(x:S))
if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
if(ttrue,x:S,l:S) -> l:S
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x:S))) -> lastbit(x:S)
zero(0) -> ttrue
zero(s(x:S)) -> ffalse
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 CONV(x:S) -> CONVITER(x:S,cons(0,nil))
 CONVITER(x:S,l:S) -> IF(zero(x:S),x:S,l:S)
 CONVITER(x:S,l:S) -> ZERO(x:S)
 HALF(s(s(x:S))) -> HALF(x:S)
 IF(ffalse,x:S,l:S) -> CONVITER(half(x:S),cons(lastbit(x:S),l:S))
 IF(ffalse,x:S,l:S) -> HALF(x:S)
 IF(ffalse,x:S,l:S) -> LASTBIT(x:S)
 LASTBIT(s(s(x:S))) -> LASTBIT(x:S)
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse

Problem 1: 

SCC Processor:
-> Pairs:
 CONV(x:S) -> CONVITER(x:S,cons(0,nil))
 CONVITER(x:S,l:S) -> IF(zero(x:S),x:S,l:S)
 CONVITER(x:S,l:S) -> ZERO(x:S)
 HALF(s(s(x:S))) -> HALF(x:S)
 IF(ffalse,x:S,l:S) -> CONVITER(half(x:S),cons(lastbit(x:S),l:S))
 IF(ffalse,x:S,l:S) -> HALF(x:S)
 IF(ffalse,x:S,l:S) -> LASTBIT(x:S)
 LASTBIT(s(s(x:S))) -> LASTBIT(x:S)
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 LASTBIT(s(s(x:S))) -> LASTBIT(x:S)
->->-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->->Cycle:
->->-> Pairs:
 HALF(s(s(x:S))) -> HALF(x:S)
->->-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->->Cycle:
->->-> Pairs:
 CONVITER(x:S,l:S) -> IF(zero(x:S),x:S,l:S)
 IF(ffalse,x:S,l:S) -> CONVITER(half(x:S),cons(lastbit(x:S),l:S))
->->-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse


The problem is decomposed in 3 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 LASTBIT(s(s(x:S))) -> LASTBIT(x:S)
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->Projection:
 pi(LASTBIT) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> Pairs:
 HALF(s(s(x:S))) -> HALF(x:S)
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->Projection:
 pi(HALF) = 1

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.3: 

Reduction Pairs Processor:
-> Pairs:
 CONVITER(x:S,l:S) -> IF(zero(x:S),x:S,l:S)
 IF(ffalse,x:S,l:S) -> CONVITER(half(x:S),cons(lastbit(x:S),l:S))
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
-> Usable rules:
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[conv](X) = 0
[conviter](X1,X2) = 0
[half](X) = 1/2.X
[if](X1,X2,X3) = 0
[lastbit](X) = 1
[zero](X) = 1/2.X + 1/2
[0] = 0
[cons](X1,X2) = 1/2.X1
[fSNonEmpty] = 0
[false] = 1
[nil] = 0
[s](X) = 2.X + 1
[true] = 0
[CONV](X) = 0
[CONVITER](X1,X2) = 2.X1 + X2 + 2
[HALF](X) = 0
[IF](X1,X2,X3) = 2.X1 + X2 + X3 + 1/2
[LASTBIT](X) = 0
[ZERO](X) = 0

Problem 1.3: 

SCC Processor:
-> Pairs:
 IF(ffalse,x:S,l:S) -> CONVITER(half(x:S),cons(lastbit(x:S),l:S))
-> Rules:
 conv(x:S) -> conviter(x:S,cons(0,nil))
 conviter(x:S,l:S) -> if(zero(x:S),x:S,l:S)
 half(0) -> 0
 half(s(0)) -> 0
 half(s(s(x:S))) -> s(half(x:S))
 if(ffalse,x:S,l:S) -> conviter(half(x:S),cons(lastbit(x:S),l:S))
 if(ttrue,x:S,l:S) -> l:S
 lastbit(0) -> 0
 lastbit(s(0)) -> s(0)
 lastbit(s(s(x:S))) -> lastbit(x:S)
 zero(0) -> ttrue
 zero(s(x:S)) -> ffalse
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.