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

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