Explaination to shorten code

 return ((aSmile && bSmile) || (!aSmile && !bSmile));

this will be true if any of the two parts is true --> it will be truye
if either both aSmile and bSmile are true (aSmile && bSmile)
or if they are both false, then (!aSmile && !bSmile));

as boolean can be either true or false this two cases cover
all cases when these two bollean will be equal to each other,
therefore it is enough to check aSmile == bSmile

And finally you can just check all four cases:

aSmil = true,  bSmile = true ---> ((aSmile && bSmile) || (!aSmile && !bSmile)) true, as aSmile && bSmile this is true
                                                    aSmile == bSmile  also true

aSmile=true  bSmile = false --> ((aSmile && bSmile) || (!aSmile && !bSmile)) false as both parts are false
                                                aSmile == bSmile  also false

aSmaile = ffalse bSmaile = true --->  the same as previous  - false
                                                               aSmile == bSmile  

aSmaile = fdalse bSmile = false --> ((aSmile && bSmile) || (!aSmile && !bSmile))  true as second part is true
                                                           aSmile == bSmile  true
   
So we see that two expressions in all cases give the same value

small correction:
And finally you can just check all four cases:

aSmil = true,  bSmile = true ---> ((aSmile && bSmile) || (!aSmile && !bSmile)) true, as aSmile && bSmile this is true
                                                    aSmile == bSmile  also true

aSmile=true  bSmile = false --> ((aSmile && bSmile) || (!aSmile && !bSmile)) false as both parts are false
                                                aSmile == bSmile  also false

aSmaile = ffalse bSmaile = true --->  the same as previous  - false
                                                               aSmile == bSmile  - false

aSmaile = fdalse bSmile = false --> ((aSmile && bSmile) || (!aSmile && !bSmile))  true as second part is true
                                                           aSmile == bSmile  true
   
So we see that two expressions in all cases give the same value

>>What is the explaination to shorten the below code from:

the second one is the optimized one in such a way that the check happens just once but there is a chance that first one execute two checks. another advantage is the second one is short and clean.

Thanks all.