/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

Problem 1: 

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

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 append(l1,l2) -> ifappend(l1,l2,is_empty(l1))
 hd(cons(x,l)) -> x
 ifappend(l1,l2,false) -> cons(hd(l1),append(tl(l1),l2))
 ifappend(l1,l2,true) -> l2
 is_empty(cons(x,l)) -> false
 is_empty(nil) -> true
 tl(cons(x,l)) -> l
-> 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,l2) -> IFAPPEND(l1,l2,is_empty(l1))
 APPEND(l1,l2) -> IS_EMPTY(l1)
 IFAPPEND(l1,l2,false) -> APPEND(tl(l1),l2)
 IFAPPEND(l1,l2,false) -> HD(l1)
 IFAPPEND(l1,l2,false) -> TL(l1)
-> Rules:
 append(l1,l2) -> ifappend(l1,l2,is_empty(l1))
 hd(cons(x,l)) -> x
 ifappend(l1,l2,false) -> cons(hd(l1),append(tl(l1),l2))
 ifappend(l1,l2,true) -> l2
 is_empty(cons(x,l)) -> false
 is_empty(nil) -> true
 tl(cons(x,l)) -> l

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.