TRS Standard pair #516963629

details

property	value
status	complete
benchmark	Ex25_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n095.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	4.36309814453 seconds
cpu usage	16.066205454
max memory	1.214959616E9

stage attributes

key	value
output-size	9887
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

Problem:
 active(f(f(X))) -> mark(c(f(g(f(X)))))
 active(c(X)) -> mark(d(X))
 active(h(X)) -> mark(c(d(X)))
 mark(f(X)) -> active(f(mark(X)))
 mark(c(X)) -> active(c(X))
 mark(g(X)) -> active(g(X))
 mark(d(X)) -> active(d(X))
 mark(h(X)) -> active(h(mark(X)))
 f(mark(X)) -> f(X)
 f(active(X)) -> f(X)
 c(mark(X)) -> c(X)
 c(active(X)) -> c(X)
 g(mark(X)) -> g(X)
 g(active(X)) -> g(X)
 d(mark(X)) -> d(X)
 d(active(X)) -> d(X)
 h(mark(X)) -> h(X)
 h(active(X)) -> h(X)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [mark](x0) = x0,
   
   [active](x0) = x0,
   
   [c](x0) = x0,
   
   [d](x0) = x0,
   
   [f](x0) = 4x0,
   
   [h](x0) = 4x0 + 1,
   
   [g](x0) = x0
  orientation:
   active(f(f(X))) = 16X >= 16X = mark(c(f(g(f(X)))))
   
   active(c(X)) = X >= X = mark(d(X))
   
   active(h(X)) = 4X + 1 >= X = mark(c(d(X)))
   
   mark(f(X)) = 4X >= 4X = active(f(mark(X)))
   
   mark(c(X)) = X >= X = active(c(X))
   
   mark(g(X)) = X >= X = active(g(X))
   
   mark(d(X)) = X >= X = active(d(X))
   
   mark(h(X)) = 4X + 1 >= 4X + 1 = active(h(mark(X)))
   
   f(mark(X)) = 4X >= 4X = f(X)
   
   f(active(X)) = 4X >= 4X = f(X)
   
   c(mark(X)) = X >= X = c(X)
   
   c(active(X)) = X >= X = c(X)
   
   g(mark(X)) = X >= X = g(X)
   
   g(active(X)) = X >= X = g(X)
   
   d(mark(X)) = X >= X = d(X)
   
   d(active(X)) = X >= X = d(X)
   
   h(mark(X)) = 4X + 1 >= 4X + 1 = h(X)
   
   h(active(X)) = 4X + 1 >= 4X + 1 = h(X)
  problem:
   active(f(f(X))) -> mark(c(f(g(f(X)))))
   active(c(X)) -> mark(d(X))
   mark(f(X)) -> active(f(mark(X)))
   mark(c(X)) -> active(c(X))
   mark(g(X)) -> active(g(X))
   mark(d(X)) -> active(d(X))
   mark(h(X)) -> active(h(mark(X)))
   f(mark(X)) -> f(X)
   f(active(X)) -> f(X)
   c(mark(X)) -> c(X)
   c(active(X)) -> c(X)
   g(mark(X)) -> g(X)
   g(active(X)) -> g(X)
   d(mark(X)) -> d(X)
   d(active(X)) -> d(X)
   h(mark(X)) -> h(X)
   h(active(X)) -> h(X)
  String Reversal Processor:

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job log

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