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


--------------------------------------------------------------------------------


YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S ys:S zs1:S zs2:S)
(RULES
le(0,y:S) -> ttrue
le(s(x:S),0) -> ffalse
le(s(x:S),s(y:S)) -> le(x:S,y:S)
split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
split(x:S,nil) -> tp2(nil,nil)
)

Problem 1: 
Valid CTRS Processor:
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)
-> The system is a deterministic 3-CTRS.

Problem 1: 

Dependency Pairs Processor:

Conditional Termination Problem 1:
-> Pairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
-> QPairs:
 Empty
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)

Conditional Termination Problem 2:
-> Pairs:
 SPLIT(x:S,cons(y:S,ys:S)) -> LE(x:S,y:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S)
 SPLIT(x:S,cons(y:S,ys:S)) -> SPLIT(x:S,ys:S)
-> QPairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)


The problem is decomposed in 2 subproblems.

Problem 1.1: 

SCC Processor:
-> Pairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
-> QPairs:
 Empty
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
-> QPairs:
 Empty
->->-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)

Problem 1.1: 

Conditional Subterm Processor:
-> Pairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
-> QPairs:
 Empty
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)
->Projection:
 pi(LE) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> QPairs:
 Empty
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

SCC Processor:
-> Pairs:
 SPLIT(x:S,cons(y:S,ys:S)) -> LE(x:S,y:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S)
 SPLIT(x:S,cons(y:S,ys:S)) -> SPLIT(x:S,ys:S)
-> QPairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 SPLIT(x:S,cons(y:S,ys:S)) -> SPLIT(x:S,ys:S)
-> QPairs:
 Empty
->->-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)

Problem 1.2: 

Conditional Subterm Processor:
-> Pairs:
 SPLIT(x:S,cons(y:S,ys:S)) -> SPLIT(x:S,ys:S)
-> QPairs:
 Empty
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)
->Projection:
 pi(SPLIT) = 2

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> QPairs:
 Empty
-> Rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 split(x:S,cons(y:S,ys:S)) -> tp2(cons(y:S,zs1:S),zs2:S) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ffalse
 split(x:S,cons(y:S,ys:S)) -> tp2(zs1:S,cons(y:S,zs2:S)) | split(x:S,ys:S) ->* tp2(zs1:S,zs2:S), le(x:S,y:S) ->* ttrue
 split(x:S,nil) -> tp2(nil,nil)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.