Q: d&(a<->d)&c|((b|b)&!d<->!c|(b<->d))

SNF with d (DNF form):

!d&!(b<->!c&!b)|d&(a&c|(c|!b))

SNF with b:

!b&(d|c&!d|!a)|b&(a&c|c)

SNF with c:

c&(!b|b&a)|!c&b|!d

SNF with a:

!a&!c|b&c|!b&d|a&c|!c&b|!d

It seems I'm unable to reach any closer to the answer : **a&b&c&d|b&!c&!d|c&d&!b**

How can I uncomplicate this process?