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HRS 58631 pair #381918963
details
property
value
status
complete
benchmark
kop12thesis_ex2.11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n038.star.cs.uiowa.edu
space
Kop_13
run statistics
property
value
solver
sol 37957
configuration
hrs
runtime (wallclock)
0.019073009491 seconds
cpu usage
0.015758346
max memory
2031616.0
stage attributes
key
value
output-size
1693
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_hrs /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES ******** General Schema criterion ******** Found constructors: cons, nil Checking type order >>OK Checking positivity of constructors >>OK Checking function dependency >>OK Checking (1) emap(F,nil) => nil (fun emap>nil) >>True Checking (2) emap(Z,cons(U,V)) => cons(Z[U],emap(%X.twice(Z,%X),V)) (fun emap>cons) (meta Z)[is acc in Z,cons(U,V)] [is acc in Z] (meta U)[is acc in Z,cons(U,V)] [is positive in cons(U,V)] [is acc in U] (fun emap=emap) subterm comparison of args w. LR LR >>False Try again using status RL Checking (1) emap(F,nil) => nil (fun emap>nil) >>True Checking (2) emap(Z,cons(U,V)) => cons(Z[U],emap(%X.twice(Z,%X),V)) (fun emap>cons) (meta Z)[is acc in Z,cons(U,V)] [is acc in Z] (meta U)[is acc in Z,cons(U,V)] [is positive in cons(U,V)] [is acc in U] (fun emap=emap) subterm comparison of args w. RL RL (fun emap>twice) (meta Z)[is acc in Z,cons(U,V)] [is acc in Z] (meta V)[is acc in Z,cons(U,V)] [is positive in cons(U,V)] [is acc in V] >>True Checking (3) twice(I,P) => I[I[P]] (meta I)[is acc in I,P] [is acc in I] (meta I)[is acc in I,P] [is acc in I] (meta P)[is acc in I,P] [is acc in P] >>True #SN! ******** Signature ******** cons : (nat,list) -> list emap : ((nat -> nat),list) -> list nil : list twice : ((nat -> nat),nat) -> nat ******** Computation Rules ******** (1) emap(F,nil) => nil (2) emap(Z,cons(U,V)) => cons(Z[U],emap(%X.twice(Z,%X),V)) (3) twice(I,P) => I[I[P]] YES
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