/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

Problem:
 div(X,e()) -> i(X)
 i(div(X,Y)) -> div(Y,X)
 div(div(X,Y),Z) -> div(Y,div(i(X),Z))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 1]  
   [i](x0) = [0 0 0]x0
             [0 0 0]  ,
   
                   [1 0 1]     [1 0 1]  
   [div](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  ,
   
         [1]
   [e] = [0]
         [0]
  orientation:
                [1 0 1]    [1]    [1 0 1]        
   div(X,e()) = [0 0 0]X + [0] >= [0 0 0]X = i(X)
                [0 0 0]    [0]    [0 0 0]        
   
                 [1 0 1]    [1 0 1]     [1 0 1]    [1 0 1]            
   i(div(X,Y)) = [0 0 0]X + [0 0 0]Y >= [0 0 0]X + [0 0 0]Y = div(Y,X)
                 [0 0 0]    [0 0 0]     [0 0 0]    [0 0 0]            
   
                     [1 0 1]    [1 0 1]    [1 0 1]     [1 0 1]    [1 0 1]    [1 0 1]                      
   div(div(X,Y),Z) = [0 0 0]X + [0 0 0]Y + [0 0 0]Z >= [0 0 0]X + [0 0 0]Y + [0 0 0]Z = div(Y,div(i(X),Z))
                     [0 0 0]    [0 0 0]    [0 0 0]     [0 0 0]    [0 0 0]    [0 0 0]                      
  problem:
   i(div(X,Y)) -> div(Y,X)
   div(div(X,Y),Z) -> div(Y,div(i(X),Z))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [i](x0) = 2x0,
    
    [div](x0, x1) = 2x0 + x1 + 1
   orientation:
    i(div(X,Y)) = 4X + 2Y + 2 >= X + 2Y + 1 = div(Y,X)
    
    div(div(X,Y),Z) = 4X + 2Y + Z + 3 >= 4X + 2Y + Z + 2 = div(Y,div(i(X),Z))
   problem:
    
   Qed