/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:
 del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
 f(true(),x,y,z) -> del(.(y,z))
 f(false(),x,y,z) -> .(x,del(.(y,z)))
 =(nil(),nil()) -> true()
 =(.(x,y),nil()) -> false()
 =(nil(),.(y,z)) -> false()
 =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))

Proof:
 Matrix Interpretation Processor: dim=2
  
  interpretation:
            [0]
   [true] = [0],
   
         [0]
   [v] = [2],
   
               [1 1]  
   [del](x0) = [0 2]x0,
   
                         [1 0]     [2 0]     [3 0]     [2 0]     [3]
   [f](x0, x1, x2, x3) = [1 0]x0 + [1 0]x1 + [3 0]x2 + [2 0]x3 + [2],
   
             [0]
   [false] = [0],
   
                 [2 0]     [1 0]     [1]
   [.](x0, x1) = [1 0]x0 + [1 0]x1 + [0],
   
         [0]
   [u] = [0],
   
                   [2 0]     [1 1]  
   [and](x0, x1) = [0 0]x0 + [0 0]x1,
   
           [0]
   [nil] = [3],
   
                 [1 0]       
   [=](x0, x1) = [0 0]x0 + x1
  orientation:
                      [3 0]    [4 0]    [2 0]    [3]    [3 0]    [4 0]    [2 0]    [3]                  
   del(.(x,.(y,z))) = [2 0]x + [4 0]y + [2 0]z + [2] >= [2 0]x + [4 0]y + [2 0]z + [2] = f(=(x,y),x,y,z)
   
                     [2 0]    [3 0]    [2 0]    [3]    [3 0]    [2 0]    [1]              
   f(true(),x,y,z) = [1 0]x + [3 0]y + [2 0]z + [2] >= [2 0]y + [2 0]z + [0] = del(.(y,z))
   
                      [2 0]    [3 0]    [2 0]    [3]    [2 0]    [3 0]    [2 0]    [2]                   
   f(false(),x,y,z) = [1 0]x + [3 0]y + [2 0]z + [2] >= [1 0]x + [3 0]y + [2 0]z + [1] = .(x,del(.(y,z)))
   
                    [0]    [0]         
   =(nil(),nil()) = [3] >= [0] = true()
   
                     [2 0]    [1 0]    [1]    [0]          
   =(.(x,y),nil()) = [0 0]x + [0 0]y + [3] >= [0] = false()
   
                     [2 0]    [1 0]    [1]    [0]          
   =(nil(),.(y,z)) = [1 0]y + [1 0]z + [0] >= [0] = false()
   
                          [2 0]    [1 0]    [2]    [2 0]    [1 0]    [2]                         
   =(.(x,y),.(u(),v())) = [0 0]x + [0 0]y + [0] >= [0 0]x + [0 0]y + [0] = and(=(x,u()),=(y,v()))
  problem:
   del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
   =(nil(),nil()) -> true()
   =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [true] = 0,
    
    [v] = 2,
    
    [del](x0) = 7x0 + 4,
    
    [f](x0, x1, x2, x3) = 4x0 + 4x1 + 4x2 + 4x3,
    
    [.](x0, x1) = 3x0 + x1,
    
    [u] = 1,
    
    [and](x0, x1) = x0 + x1,
    
    [nil] = 4,
    
    [=](x0, x1) = 4x0 + 2x1
   orientation:
    del(.(x,.(y,z))) = 21x + 21y + 7z + 4 >= 20x + 12y + 4z = f(=(x,y),x,y,z)
    
    =(nil(),nil()) = 24 >= 0 = true()
    
    =(.(x,y),.(u(),v())) = 12x + 4y + 10 >= 4x + 4y + 6 = and(=(x,u()),=(y,v()))
   problem:
    
   Qed