awk + one line awk syntax to print only once the second field if "true" word matched

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I don't know how to do this with
awk
But my target is to create awk one line syntax in order to print the
second field ($2) if
all first field ($1) are true

for
example

I have the
file:

 true
            my_name_is_lenon
 true my_name_is_lenon
 true
            my_name_is_lenon
 false my_name_is_lenon
 true
            my_name_is_lenon


 false my_dog_is_fat
 true
            my_dog_is_fat
 true my_dog_is_fat

 true
            I_am_very_tall
 true I_am_very_tall
 true
            I_am_very_tall

 true my_ball_is_fine
 true
            my_ball_is_fine

the
awk will print only the
following:

            I_am_very_tall

            my_ball_is_fine

Because
I_am_very_tall , my_ball_is_fine get true on the first field without
false

my_dog_is_fat ,
my_name_is_lenon not printed because the false word on the first
field

The rule is to print the second field if
no false word on the first field! (Of all the same sentence on the second
field)

lidia

Answer

Here is one
way:

awk '{if ($1 == "false")
            {array[$2] = $1} else if (array[$2] != "false") array[$2] = $1} END {for (i in array) if
            (array[i] == "true") print i}'
            inputfile

Edit:

Here
it is on multiple lines:

awk '{if
            ($1 == "false") 
 {array[$2] = $1} 
 else if (array[$2] != "false")
            
 array[$2] = $1} 
 END {for (i in array)

 if
            (array[i] == "true") 
 print i}
'
            inputfile