TRS Standard pair #615296945

details

property	value
status	complete
benchmark	2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	Secret_06_TRS

run statistics

property	value
solver	ttt2-1.20
configuration	ttt2
runtime (wallclock)	0.668736934662 seconds
cpu usage	1.44146548
max memory	2.14638592E8

stage attributes

key	value
output-size	3438
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:
 c(c(c(b(x)))) -> a(1(),b(c(x)))
 b(c(b(c(x)))) -> a(0(),a(1(),x))
 a(0(),x) -> c(c(x))
 a(1(),x) -> c(b(x))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
         [3]
   [0] = [1],
   
             [1 1]     [1]
   [c](x0) = [1 0]x0 + [1],
   
                 [1 1]     [2 1]     [0]
   [a](x0, x1) = [0 1]x0 + [1 1]x1 + [1],
   
             [1 1]  
   [b](x0) = [1 0]x0,
   
         [1]
   [1] = [0]
  orientation:
                   [5 3]    [6]    [5 3]    [6]                 
   c(c(c(b(x)))) = [3 2]x + [4] >= [3 2]x + [4] = a(1(),b(c(x)))
   
                   [5 3]    [7]    [5 3]    [7]                  
   b(c(b(c(x)))) = [3 2]x + [4] >= [3 2]x + [4] = a(0(),a(1(),x))
   
              [2 1]    [4]    [2 1]    [3]          
   a(0(),x) = [1 1]x + [2] >= [1 1]x + [2] = c(c(x))
   
              [2 1]    [1]    [2 1]    [1]          
   a(1(),x) = [1 1]x + [1] >= [1 1]x + [1] = c(b(x))
  problem:
   c(c(c(b(x)))) -> a(1(),b(c(x)))
   b(c(b(c(x)))) -> a(0(),a(1(),x))
   a(1(),x) -> c(b(x))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [0] = 0,
    
    [c](x0) = x0 + 6,
    
    [a](x0, x1) = 2x0 + x1 + 4,
    
    [b](x0) = x0 + 2,
    
    [1] = 4
   orientation:
    c(c(c(b(x)))) = x + 20 >= x + 20 = a(1(),b(c(x)))
    
    b(c(b(c(x)))) = x + 16 >= x + 16 = a(0(),a(1(),x))
    
    a(1(),x) = x + 12 >= x + 8 = c(b(x))
   problem:
    c(c(c(b(x)))) -> a(1(),b(c(x)))
    b(c(b(c(x)))) -> a(0(),a(1(),x))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
           [1]
     [0] = [0]
           [0],
     
               [1 1 0]     [0]
     [c](x0) = [0 0 1]x0 + [0]
               [0 0 0]     [1],
     
                        [1 0 0]  
     [a](x0, x1) = x0 + [0 0 0]x1
                        [0 0 0]  ,
     
               [1 1 0]  
     [b](x0) = [0 0 1]x0
               [0 0 0]  ,
     
           [0]
     [1] = [1]
           [1]
    orientation:
                     [1 1 1]    [1]    [1 1 1]    [0]                 
     c(c(c(b(x)))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a(1(),b(c(x)))
                     [0 0 0]    [1]    [0 0 0]    [1]                 
     
                     [1 1 1]    [1]    [1 0 0]    [1]                  
     b(c(b(c(x)))) = [0 0 0]x + [1] >= [0 0 0]x + [0] = a(0(),a(1(),x))
                     [0 0 0]    [0]    [0 0 0]    [0]                  
    problem:

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