SRS Stand 10685 pair #381717561

details

property	value
status	complete
benchmark	x05.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n046.star.cs.uiowa.edu
space	Secret_07_SRS

run statistics

property	value
solver	ttt2-1.17+nonreach
configuration	ttt2-1.17+nonreach
runtime (wallclock)	23.3273210526 seconds
cpu usage	91.860506768
max memory	9.8146304E8

stage attributes

key	value
output-size	5165
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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


YES

Problem:
 a(a(x1)) -> b(b(x1))
 c(c(b(x1))) -> d(c(a(x1)))
 a(x1) -> d(c(c(x1)))
 c(d(x1)) -> b(c(x1))

Proof:
 String Reversal Processor:
  a(a(x1)) -> b(b(x1))
  b(c(c(x1))) -> a(c(d(x1)))
  a(x1) -> c(c(d(x1)))
  d(c(x1)) -> c(b(x1))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [0]
    [d](x0) = [0 0 1]x0 + [0]
              [0 0 0]     [1],
    
              [1 0 0]  
    [c](x0) = [0 0 0]x0
              [0 0 1]  ,
    
              [1 0 0]     [0]
    [b](x0) = [1 0 0]x0 + [1]
              [0 0 0]     [1],
    
              [1 1 0]     [0]
    [a](x0) = [1 1 0]x0 + [1]
              [0 1 0]     [1]
   orientation:
               [2 2 0]     [1]    [1 0 0]     [0]           
    a(a(x1)) = [2 2 0]x1 + [2] >= [1 0 0]x1 + [1] = b(b(x1))
               [1 1 0]     [2]    [0 0 0]     [1]           
    
                  [1 0 0]     [0]    [1 0 0]     [0]              
    b(c(c(x1))) = [1 0 0]x1 + [1] >= [1 0 0]x1 + [1] = a(c(d(x1)))
                  [0 0 0]     [1]    [0 0 0]     [1]              
    
            [1 1 0]     [0]    [1 0 0]     [0]              
    a(x1) = [1 1 0]x1 + [1] >= [0 0 0]x1 + [0] = c(c(d(x1)))
            [0 1 0]     [1]    [0 0 0]     [1]              
    
               [1 0 0]     [0]    [1 0 0]     [0]           
    d(c(x1)) = [0 0 1]x1 + [0] >= [0 0 0]x1 + [0] = c(b(x1))
               [0 0 0]     [1]    [0 0 0]     [1]           
   problem:
    b(c(c(x1))) -> a(c(d(x1)))
    a(x1) -> c(c(d(x1)))
    d(c(x1)) -> c(b(x1))
   DP Processor:
    DPs:
     b#(c(c(x1))) -> d#(x1)
     b#(c(c(x1))) -> a#(c(d(x1)))
     a#(x1) -> d#(x1)
     d#(c(x1)) -> b#(x1)
    TRS:
     b(c(c(x1))) -> a(c(d(x1)))
     a(x1) -> c(c(d(x1)))
     d(c(x1)) -> c(b(x1))
    TDG Processor:
     DPs:
      b#(c(c(x1))) -> d#(x1)
      b#(c(c(x1))) -> a#(c(d(x1)))
      a#(x1) -> d#(x1)
      d#(c(x1)) -> b#(x1)
     TRS:
      b(c(c(x1))) -> a(c(d(x1)))
      a(x1) -> c(c(d(x1)))
      d(c(x1)) -> c(b(x1))
     graph:
      a#(x1) -> d#(x1) -> d#(c(x1)) -> b#(x1)
      d#(c(x1)) -> b#(x1) -> b#(c(c(x1))) -> a#(c(d(x1)))
      d#(c(x1)) -> b#(x1) -> b#(c(c(x1))) -> d#(x1)
      b#(c(c(x1))) -> a#(c(d(x1))) -> a#(x1) -> d#(x1)
      b#(c(c(x1))) -> d#(x1) -> d#(c(x1)) -> b#(x1)
     Matrix Interpretation Processor: dim=4
      
      interpretation:
       [a#](x0) = [0 1 0 0]x0,
       
       [d#](x0) = [0 1 0 0]x0,
       
       [b#](x0) = [0 0 0 1]x0,
       
                 [0 1 0 0]     [1]
                 [0 1 1 0]     [0]
       [d](x0) = [0 0 0 0]x0 + [0]
                 [0 1 0 0]     [0],
       
                 [0 1 0 0]     [1]
                 [0 0 0 1]     [0]

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