gistfile1.txt

* Question 2
** a

   No, it does not follow that that :a and :b are different. (Note: I've
   attempted to use Unicode for the notation, which is dependant on the system
   fonts to display as intended. Please let me know if it seems weird. I think I
   might be slightly ‘off’ on some of the symbols, but I think it will be
   sufficently clear to understand what I mean.)

   { Place ⊑ ⊤,
     Place ⊑ ∃name,
     Place ⊑ ∃position.Point,
     Point ⊑ ⊤,
     Point ⊑ ∃lat.double,
     Point ⊑ ∃long.double,
     Connection ⊑ ⊤,
     Connection ⊑ ∃endpoint:Place,
     Connection ⊑ ∃distance.double,

     Connection(:x),
     Place(:a),
     Place(:b),

     endpoint(:x, :a),
     endpoint(:x, :b)}

     ∆¹ = {1,2,3,4,5,6,7}
     Connection¹ = {1}
     Place¹ = {2}
     Point¹ = {3}
     
     (I'm not sure if these (string, double) should be included or not)

     string¹ = {4}           
     double¹ = {5,6,7}

     name¹ = {(2,4)}
     position¹ = {(2,3)}
     lat¹ = {(3,5)}
     long¹ = {(3,6)}
     
     endpoint¹ = {(1,2),(1,2)}
     distance¹ = {(1,7)}
     

     :x¹ = 1
     :a¹ = 2
     :b¹ = 2

     Note: I'm not sure how I define cardinality in an interpretation, but I
     feel I it should be ‘said‘ that Place has at least one name and exactly
     one position, Point has exactly one lat and lang and that Connection has
     exactly one distance and two endpoints.

     - I satisfies Place ⊑ ⊤, since {2} ⊆ {1,2,3,4,5,6,7}
     - I satisfies Place ⊑ ∃name, since Place¹ = {2}, string¹ = {4} and (2,4) ⊆ name¹ = {(2,4)}
     - I satisfies Place ⊑ ∃position.Point, since Place¹ = {2}, Point¹ = {3} and (2,3) ⊆ position¹ = {(2,3)}
     - I satisfies Point ⊑ ⊤, since {3} ⊆ {1,2,3,4,5,6,7}
     - I satisfies Point ⊑ ∃lat.double, since Point¹ = {3}, double¹ = {5,6,7} and (3,5) ⊆ lat¹ = {(3,5)}
     - I satisfies Point ⊑ ∃long.double, since Point¹ = {3}, double¹ = {5,6,7} and (3,6) ⊆ long¹ = {(3,6)}
     - I satisfies Connection ⊑ ⊤, since {1} ⊆ {1,2,3,4,5,6,7}
     - I satisfies Connection ⊑ ∃endpoint:Place, since Connection¹ = {1}, Place¹ = {2} and (1,2) ⊆ endpoint¹ = {(1,2),(1,2)}
     - I satisfies Connection ⊑ ∃distance.double, since Connection¹ = {1}, double¹ = {5,6,7} and (1,7) ⊆ distance¹ = {(3,6)}

     - I satisfies Connection(:x), since :x¹ = 1 and 1 ∈ Connection¹ = {1}
     - I satisfies Place(:a), since :a¹ = 2 and 2 ∈ Place¹ = {2}
     - I satisfies Place(:b), since :b¹ = 2 and 2 ∈ Place¹ = {2}
     - I satisfies endpoint(:x, :a), since :x¹ = 1, :a¹ = 2 and (1,2) ∈ endpoint¹ = {(1,2),(1,2)}
     - I satisfies endpoint(:x, :b), since :x¹ = 1, :b¹ = 2 and (1,2) ∈ endpoint¹ = {(1,2),(1,2)}

       Since :a = 2, :b = 2 and 2 = 2, it does not follow that :a and :b are different.


** b

   No, it does not follow that :a and :b are the same.

   This model is similar to the one above.

   ∆¹ = {1,2,3,4,5,6,7}
   Connection¹ = {1}
   Place¹ = {2,8}
   Point¹ = {3,9}

   string¹ = {4,10}           
   double¹ = {5,6,7,11,12}
   
   name¹ = {(2,4),(8,10)}
   position¹ = {(2,3),(8,9)}
   lat¹ = {(3,5),(9,11)}
   long¹ = {(3,6),(9,12)}
     
   endpoint¹ = {(1,2),(1,8)}
   distance¹ = {(1,7)}

   :x¹ = 1
   :a¹ = 2
   :b¹ = 8

   - I satisfies Place ⊑ ⊤, since {2,8} ⊆ {1,2,3,4,5,6,7,8,9,10,11,12}
   - I satisfies Place ⊑ ∃name, since Place¹ = {2,8}, string¹ = {4,10} and {(2,4),(8,10)} ⊆ name¹ = {(2,4),(8,10)}
   - I satisfies Place ⊑ ∃position.Point, since Place¹ = {2,8}, Point¹ = {3,9} and {(2,3),(8,9)} ⊆ position¹ = {(2,3),(8,9)}
   - I satisfies Point ⊑ ⊤, since {3,9} ⊆ {1,2,3,4,5,6,7,8,9,10,11,12}
   - I satisfies Point ⊑ ∃lat.double, since Point¹ = {3,9}, double¹ = {5,6,7,11,12} and {(3,5),(9,11)} ⊆ lat¹ = {(3,5),(9,11)}
   - I satisfies Point ⊑ ∃long.double, since Point¹ = {3,9}, double¹ = {5,6,7,11,12} and {(3,6),(9,12)} ⊆ long¹ = {(3,6),(9,12)}
   - I satisfies Connection ⊑ ⊤, since {1} ⊆ {1,2,3,4,5,6,7,8,9,10,11,12}
   - I satisfies Connection ⊑ ∃endpoint:Place, since Connection¹ = {1}, Place¹ = {2,8} and {(1,2),(1,8)} ⊆ endpoint¹ = {(1,2),(1,8)}
   - I satisfies Connection ⊑ ∃distance.double, since Connection¹ = {1}, double¹ = {5,6,7,11,12} and (1,7) ⊆ distance¹ = {(3,6)}

   - I satisfies Connection(:x), since :x¹ = 1 and 1 ∈ Connection¹ = {1}
   - I satisfies Place(:a), since :a¹ = 2 and 2 ∈ Place¹ = {2,8}
   - I satisfies Place(:b), since :b¹ = 8 and 8 ∈ Place¹ = {2,8}
   - I satisfies endpoint(:x, :a), since :x¹ = 1, :a¹ = 2 and (1,2) ∈ endpoint¹ = {(1,2),(1,8)}
   - I satisfies endpoint(:x, :b), since :x¹ = 1, :b¹ = 8 and (1,8) ∈ endpoint¹ = {(1,2),(1,8)}

       Since :a = 2, :b = 8 and 2 ≠ 8, it does not follow that :a and :b are the same.

** c

   If we want :a and :b to be different:
     :a owl:differentFrom :b .

   If we want :a and :b to be the same:
     :a owl:sameAs :b .