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TRS Standard pair #516963879
details
property
value
status
complete
benchmark
Liveness6.1.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n063.star.cs.uiowa.edu
space
AProVE_04
run statistics
property
value
solver
ttt2-1.20
configuration
ttt2
runtime (wallclock)
0.682222127914 seconds
cpu usage
1.355803985
max memory
2.51715584E8
stage attributes
key
value
output-size
8740
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: top(free(x)) -> top(check(new(x))) new(free(x)) -> free(new(x)) old(free(x)) -> free(old(x)) new(serve()) -> free(serve()) old(serve()) -> free(serve()) check(free(x)) -> free(check(x)) check(new(x)) -> new(check(x)) check(old(x)) -> old(check(x)) check(old(x)) -> old(x) Proof: Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [1] [old](x0) = [1 0 0]x0 + [1] [1 0 0] [1], [1 0 0] [0] [top](x0) = [0 0 0]x0 + [0] [0 0 1] [1], [1 0 0] [check](x0) = [0 0 1]x0 [0 1 0] , [0] [serve] = [0] [0], [1 0 0] [free](x0) = [1 0 0]x0 [1 0 0] , [1 0 0] [new](x0) = [1 0 0]x0 [1 0 0] orientation: [1 0 0] [0] [1 0 0] [0] top(free(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = top(check(new(x))) [1 0 0] [1] [1 0 0] [1] [1 0 0] [1 0 0] new(free(x)) = [1 0 0]x >= [1 0 0]x = free(new(x)) [1 0 0] [1 0 0] [1 0 0] [1] [1 0 0] [1] old(free(x)) = [1 0 0]x + [1] >= [1 0 0]x + [1] = free(old(x)) [1 0 0] [1] [1 0 0] [1] [0] [0] new(serve()) = [0] >= [0] = free(serve()) [0] [0] [1] [0] old(serve()) = [1] >= [0] = free(serve()) [1] [0] [1 0 0] [1 0 0] check(free(x)) = [1 0 0]x >= [1 0 0]x = free(check(x)) [1 0 0] [1 0 0] [1 0 0] [1 0 0] check(new(x)) = [1 0 0]x >= [1 0 0]x = new(check(x)) [1 0 0] [1 0 0] [1 0 0] [1] [1 0 0] [1] check(old(x)) = [1 0 0]x + [1] >= [1 0 0]x + [1] = old(check(x)) [1 0 0] [1] [1 0 0] [1] [1 0 0] [1] [1 0 0] [1] check(old(x)) = [1 0 0]x + [1] >= [1 0 0]x + [1] = old(x) [1 0 0] [1] [1 0 0] [1] problem: top(free(x)) -> top(check(new(x))) new(free(x)) -> free(new(x)) old(free(x)) -> free(old(x)) new(serve()) -> free(serve()) check(free(x)) -> free(check(x)) check(new(x)) -> new(check(x)) check(old(x)) -> old(check(x)) check(old(x)) -> old(x) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [1] [old](x0) = [0 1 0]x0 + [0] [0 0 0] [0], [1 1 1] [top](x0) = [1 0 1]x0
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