/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/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=1
  
  interpretation:
   [f](x0, x1) = x0 + x1 + 1,
   
   [h](x0) = x0,
   
   [minus](x0) = 4x0 + 3
  orientation:
   minus(minus(x)) = 16x + 15 >= x = x
   
   minus(h(x)) = 4x + 3 >= 4x + 3 = h(minus(x))
   
   minus(f(x,y)) = 4x + 4y + 7 >= 4x + 4y + 7 = f(minus(y),minus(x))
  problem:
   minus(h(x)) -> h(minus(x))
   minus(f(x,y)) -> f(minus(y),minus(x))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
                  [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],
    
              [1 0 0]     [0]
    [h](x0) = [0 0 0]x0 + [1]
              [0 1 1]     [0],
    
                  [1 1 1]  
    [minus](x0) = [0 1 0]x0
                  [1 0 1]  
   orientation:
                  [1 1 1]    [1]    [1 1 1]    [0]              
    minus(h(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = h(minus(x))
                  [1 1 1]    [0]    [1 1 1]    [0]              
    
                    [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