Sets theory and Mathematical logic per K. Polyakov && Bitwise2 witchcraft in action, samples of tasks 18 EGE 2018 at VK's newswire in 10/2017

Consider the most recent tasks of type 18 submitted to VK's newswire on 09/10/17

   Follow http://kpolyakov.spb.ru/download/inf-2015-10.pdf

   ¬A => ¬(¬P => Q) =1
   ¬A => ¬(P v Q) = 1
   ¬A => (¬P^¬Q) =1 
     A v  (¬P^¬Q) =1 

Conditions of Task 1 from link above are satisfied, so
 
     A(min) = ¬(¬P^¬Q) 
     A(min) =(P v Q) = [4;37]

Answer : 33 

  
   ¬A => ¬(¬P => Q) =1
   ¬A => ¬(P v Q) = 1
   ¬A => (¬P^¬Q) =1 
     A v  (¬P^¬Q) =1 

Conditions of Task 1 from link above are satisfied, so
 
     A(min) = ¬(¬P^¬Q) 
     A(min) =(P v Q) = [7;56]

Answer : 49

Basic concepts explained in
http://kpolyakov.spb.ru/download/inf-2015-10.pdf
to automate solution the tasks 18 dealing with segments

  

  How to stop this headache ?

  

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Bitwise2 witchcraft http://kpolyakov.spb.ru/download/bitwise2.pdf

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  A ^ (¬Z(35) => ¬Z(52)) = 0

¬A + Z(35) + ¬Z(52) = 1
¬(AZ(52)) + Z(35) = 1
AZ(52) => Z(35) =1
Z(52 or A) => Z(35) =1

35 = 100011
52 = 110100
===========
A =  000011

A(min) =3

    ¬A => (Z(36) => ¬Z(6)) = 1
      A + ¬Z(36) + ¬Z(6) = 1
   ¬(Z(36)Z(6)) + A = 1
     (Z(36)Z(6)) => A = 1
     Z(36 or 6) => A = 1

    36=100100
    v
    6 =000110
    =============
   A(max) = 100110 = 38 (decimal)
   

   ¬Z(77) => (Z(12) => ¬A) = 1
     Z(77) + ¬Z(12) + ¬A = 1
   ¬(Z(12)A) + Z(77) = 1
     Z(12)A => Z(77) = 1
     Z(12 or A) => Z(77) = 1

77 = 1001101
12 = 0001100
=========
A(min)  = 1000001(binary) = 65 (decimal)
  
  Automate solution the task 18 dealing with segments
    

   Follow http://kpolyakov.spb.ru/download/inf-2015-10.pdf 

   (P => Q)^A = 0

   (¬P v Q)^A = 0
¬((¬P v Q)^A) = 1
   (P ^¬Q) v ¬A = 1

  Conditions of Task 2 from link above are satisfied, so

   A(max) = (P ^¬Q) = [44;48]