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


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YES

Problem:
 minus(minus(x)) -> x
 minus(h(x)) -> h(minus(x))
 minus(f(x,y)) -> f(minus(y),minus(x))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 1]     [0]
   [h](x0) = [0 0 1]x0 + [1]
             [0 1 0]     [1],
   
                 [1 0 1]  
   [minus](x0) = [0 0 1]x0
                 [0 1 0]  ,
   
                 [1 0 0]     [1 0 1]  
   [f](x0, x1) = [0 0 0]x0 + [0 1 1]x1
                 [0 1 1]     [0 0 0]  
  orientation:
                     [1 1 1]          
   minus(minus(x)) = [0 1 0]x >= x = x
                     [0 0 1]          
   
                 [1 1 1]    [1]    [1 1 1]    [0]              
   minus(h(x)) = [0 1 0]x + [1] >= [0 1 0]x + [1] = h(minus(x))
                 [0 0 1]    [1]    [0 0 1]    [1]              
   
                   [1 1 1]    [1 0 1]     [1 1 1]    [1 0 1]                        
   minus(f(x,y)) = [0 1 1]x + [0 0 0]y >= [0 1 1]x + [0 0 0]y = f(minus(y),minus(x))
                   [0 0 0]    [0 1 1]     [0 0 0]    [0 1 1]                        
  problem:
   minus(minus(x)) -> x
   minus(f(x,y)) -> f(minus(y),minus(x))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [minus](x0) = 2x0 + 1,
    
    [f](x0, x1) = 4x0 + 4x1 + 7
   orientation:
    minus(minus(x)) = 4x + 3 >= x = x
    
    minus(f(x,y)) = 8x + 8y + 15 >= 8x + 8y + 15 = f(minus(y),minus(x))
   problem:
    minus(f(x,y)) -> f(minus(y),minus(x))
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                   [1 1 1]  
     [minus](x0) = [0 1 0]x0
                   [1 0 1]  ,
     
                   [1 0 0]     [1 0 0]     [0]
     [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 + [1]
                   [0 1 1]     [0 1 1]     [0]
    orientation:
                     [1 1 1]    [1 1 1]    [1]    [1 1 1]    [1 1 1]    [0]                       
     minus(f(x,y)) = [0 0 0]x + [0 0 0]y + [1] >= [0 0 0]x + [0 0 0]y + [1] = f(minus(y),minus(x))
                     [1 1 1]    [1 1 1]    [0]    [1 1 1]    [1 1 1]    [0]                       
    problem:
     
    Qed