Morphological Filters

• Translation: (A)_z= {a+z| (a within A)}
• Reflection: (~B~)={w|w=-b (b within B)}
• Dilation: (A_dilation_B)={z|(~B~)_z (intersection) A != [EMPTY_GROUP]}
• Erosion: (A_erosion_B)={z|(B)_z (contained in) A}
• Opening: (A_opening_B)=(A_erosion_B)_dilation_B
• Closing: (A_closing_B)=(A_dilation_B)_erosion_B

We want to show that:
A_closing_B (contains) A

We assume that there is element a in A.
Lets say there is an element x in A_closing_B.
That means that x is in (A_dilation_B)_erosion_B.
And for some x’ in (A_dilation_B) we can say that its in {z|(~B~)_z (intersection) A != [EMPTY_GROUP]}, so for every b in B exists some a in A that satisfies : x’+b=a.

Lets call the group of (A_dilation_B) group C.
So x is in C_erosion_B (and x’ is in C), that means x in {z|(B)_z (contained in) C} and from that we get that for every b in B exists some c in C that
satisfies: -b+x = c.

We know that x’ is in C so we can say that c=x’ so from both of the bold formulas we get that -b+x =a-b which means x=a.