/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:
 a(b(x1)) -> b(c(a(x1)))
 b(c(x1)) -> c(b(b(x1)))
 b(a(x1)) -> a(c(b(x1)))

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(c(b(x1)))
  c(b(x1)) -> b(b(c(x1)))
  a(b(x1)) -> b(c(a(x1)))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
              [1 0 0]     [1]
    [a](x0) = [0 0 0]x0 + [0]
              [0 0 1]     [1],
    
              [1 0 1]  
    [b](x0) = [0 0 0]x0
              [0 0 1]  ,
    
              [1 0 0]     [0]
    [c](x0) = [1 0 0]x0 + [1]
              [0 0 0]     [0]
   orientation:
               [1 0 1]     [2]    [1 0 1]     [1]              
    b(a(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = a(c(b(x1)))
               [0 0 1]     [1]    [0 0 0]     [1]              
    
               [1 0 1]     [0]    [1 0 0]                
    c(b(x1)) = [1 0 1]x1 + [1] >= [0 0 0]x1 = b(b(c(x1)))
               [0 0 0]     [0]    [0 0 0]                
    
               [1 0 1]     [1]    [1 0 0]     [1]              
    a(b(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = b(c(a(x1)))
               [0 0 1]     [1]    [0 0 0]     [0]              
   problem:
    c(b(x1)) -> b(b(c(x1)))
    a(b(x1)) -> b(c(a(x1)))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(c) = w(a) = w(b) = 0
    status function:
     st(c) = st(a) = st(b) = [0]
    precedence:
     a > c > b
    problem:
     
    Qed