RockMasterrans
Answered

[P=>Pvq]=>(r^q') en sade hali



Sagot :

Superbayrans

[p'v(pvq)' ]v (r'^q')

[p'v(p'vq') ]'v(r'^q')

(p'vq')'v(r'^q')

(p^q)v(r'^q')

(p^r')v(q'^q)

(p^r')v0

p^r' 

ama cok emın deılım

buyuk ıhtımalle bu ama

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