SRS Stand 10685 pair #381717210

details

property	value
status	complete
benchmark	hom01.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	Trafo_06

run statistics

property	value
solver	ttt2-1.17+nonreach
configuration	ttt2-1.17+nonreach
runtime (wallclock)	2.65995502472 seconds
cpu usage	9.352951708
max memory	1.001480192E9

stage attributes

key	value
output-size	2437
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES

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

Proof:
 String Reversal Processor:
  a(d(b(d(b(a(a(x1))))))) -> d(b(a(a(c(a(a(x1)))))))
  c(a(a(x1))) -> a(a(c(c(x1))))
  c(c(c(x1))) -> d(b(c(d(b(x1)))))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]  
    [c](x0) = [0 1 0]x0
              [0 0 0]  ,
    
              [1 0 0]  
    [b](x0) = [0 1 0]x0
              [0 1 1]  ,
    
              [1 0 0]  
    [d](x0) = [0 0 0]x0
              [0 1 1]  ,
    
              [1 0 1]     [0]
    [a](x0) = [0 0 0]x0 + [1]
              [0 0 1]     [0]
   orientation:
                              [1 0 3]     [2]    [1 0 2]     [0]                          
    a(d(b(d(b(a(a(x1))))))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [0] = d(b(a(a(c(a(a(x1)))))))
                              [0 0 1]     [2]    [0 0 0]     [2]                          
    
                  [1 0 2]     [0]    [1 0 0]     [0]                 
    c(a(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(a(c(c(x1))))
                  [0 0 0]     [0]    [0 0 0]     [0]                 
    
                  [1 0 0]      [1 0 0]                      
    c(c(c(x1))) = [0 1 0]x1 >= [0 0 0]x1 = d(b(c(d(b(x1)))))
                  [0 0 0]      [0 0 0]                      
   problem:
    c(a(a(x1))) -> a(a(c(c(x1))))
    c(c(c(x1))) -> d(b(c(d(b(x1)))))
   String Reversal Processor:
    a(a(c(x1))) -> c(c(a(a(x1))))
    c(c(c(x1))) -> b(d(c(b(d(x1)))))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {6,1}
      transitions:
       b1(27) -> 28*
       b1(24) -> 25*
       b1(21) -> 22*
       b1(18) -> 19*
       d1(20) -> 21*
       d1(17) -> 18*
       d1(26) -> 27*
       d1(23) -> 24*
       c1(25) -> 26*
       c1(19) -> 20*
       f40() -> 2*
       c0(5) -> 1*
       c0(4) -> 5*
       c0(8) -> 9*
       a0(2) -> 3*
       a0(3) -> 4*
       b0(10) -> 6*
       b0(7) -> 8*
       d0(2) -> 7*
       d0(9) -> 10*
       1 -> 3,4
       4 -> 23*
       5 -> 17*
       22 -> 1,4
       28 -> 5*
     problem:
      
     Qed

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