its simple just go step by step and break it:-
S1 can be broken into two components :
∃x∃y[A(x)⋀A(y)⋀(x≠y) and ∀z(A(z)→(x=z⋁y=z)]
statement 1 : it says there exist some x and y such that x and y are apples and they are distinct or it also means we have atleast two different apples.
statement 2 : for all values of z if z is an apple than either x (=z) or y(=z) means if z is an apple its either x or y.
conclusion : there exist two apples x and y which are different and if there exists any other apple rather these two it has to be either x or y.
means there are exactly two apples.
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similarly S2 can be broken down into:
∃x∃y[A(x)⋀A(y)⋀x≠y] and ~∃x∃y∃z[A(x)⋀A(y)∧A(z)⋀x≠y∧y≠z⋀z≠x]
statement 1 : similar to statement 1 of S1.
statement 2: it say there exist some apples x , y, z such that if these are apples than they need to be distinct as x≠y∧y≠z⋀z≠x or you can say that it means we have atleast 3 different apples.
conclusion : on combining statement 1 and statement 2 we can say that
= atleast 2 apples and ~atleast 3 apples.
=atleast 2 apples - atleast 3 apples.
=exactly 2 apples.
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so in both the above cases : we can say that S1 and S2 are same in all possible ways.
therefore S1 <--> S2. Finally concluded...
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