4+1 == 3+2 [enter] returns true
(5==5) [enter] returns true
((e^(π*i)+1)==0) [enter] returns true
true == true [enter] returns true
(5==5) == ((e^(π*i)+1)==0) [enter] returns false. Why?
(5==5) == ((e^(π*i)+1)==0) [enter]
entry(1) [enter] returns (((5==5)) and ((5==(exp(pi*i)+1))) and (((exp(pi*i)+1)==0))) ?
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