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


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


YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 0]  
   [b](x0) = [0 1 0]x0
             [0 1 0]  ,
   
             [1 0 1]     [0]
   [a](x0) = [0 1 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]  
   [c](x0) = [0 0 0]x0
             [0 0 1]  
  orientation:
                    [1 1 1]     [1]    [1 1 0]     [0]                 
   a(b(b(a(x1)))) = [0 1 0]x1 + [2] >= [0 0 0]x1 + [1] = a(c(a(b(x1))))
                    [0 0 0]     [0]    [0 0 0]     [0]                 
   
              [1 0 1]     [0]    [1 0 1]                
   a(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 = c(c(a(x1)))
              [0 0 0]     [0]    [0 0 0]                
   
                 [1 0 0]      [1 0 0]                
   c(c(c(x1))) = [0 0 0]x1 >= [0 0 0]x1 = b(c(b(x1)))
                 [0 0 1]      [0 0 0]                
  problem:
   a(c(x1)) -> c(c(a(x1)))
   c(c(c(x1))) -> b(c(b(x1)))
  String Reversal Processor:
   c(a(x1)) -> a(c(c(x1)))
   c(c(c(x1))) -> b(c(b(x1)))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(a) = 4
     w(c) = w(b) = 0
    status function:
     st(c) = st(b) = st(a) = [0]
    precedence:
     c > b ~ a
    problem:
     
    Qed