= (xx 2xx 4 v x3 v x3xx 4) (x2xx 4 v xx 3x4 v x3xx 4) = (xx 2xx 4 v x3) (x2xx 4 v xx 3x4 v x3xx 4) = (xx 2xx 4 v x3) (x2xx 4 v xx 3x4 v x3xx 4) v (xx 2xx 4 v x3) (x2xx 4 v xx 3x4 v x3xx 4) = (xx 2xx 4 x3) (x2xx 4 v xx 3x4 v x3xx 4) v (xx 2xx 4 v x3) (x2xx 4 xx 3x4 x3xx 4) = ((x2 v x4) x3) (x2xx 4 v xx 3x4 v x3xx 4) v (xx 2xx 4 v x3) (xx 2 v x4) (x3 v xx 4) (xx 3 v x4) = x2x3xx 4 v (xx 2xx 4 v xx 2x3 v x3x4) (x3x4 v xx 3xx 4) = x2x3xx 4 v xx 2xx 3xx 4 v x3x4 10.2.x2 järgi: f = f(x1,0,x3,x4) f(x1,1,x3,x4) = (xx 11xx 4 v xx 1x3 v x10xx 4 v x1xx 3x4 v x3xx 4) (xx 10xx 4 v xx 1x3 v x11xx 4 v x1xx 3x4 v x3xx 4) = (xx 1xx 4 v xx 1x3 v x1xx 3x4 v x3xx 4) (xx 1x3 v x1xx 4 v x1xx 3x4 v x3xx 4) = (xx 1xx 4 v xx 1x3 v x1xx 3x4 v x3xx 4) (xx 1x3 v x1xx 4 v x1xx 3x4 v x3xx 4) v (xx 1xx 4 v xx 1x3 v x1xx 3x4 v x3xx 4) (xx 1x3 v x1xx 4 v x1xx 3x4 v x3xx 4) = (xx 1xx 4 xx 1x3 x1xx 3x4 x3xx 4)
Password: 1234 Password: 1234 Caller ID Name: Homer Caller ID Name: Bart <150> Homer <250> Bart, . IAX 2 PBX2 PBX3. PBX2 . : FreePBX menu => Add Trunk => Add IAX2 Trunk Outgoing Settings [300peer] host=192.168.252.43 username=200user secret=1234 qualify=yes trunk=yes type=peer Incoming Settings [300user] secret=1234 type=user context=fromtrunk FreePBX menu => Outbound Routes [300dial3xx] [3xx] [IAX2/300peer] PBX3. https ://192.168.252.43 , . AskoziaPBX =>Accaunts => Providers, : Name: 200user Username: 300user Password: 1234 Host: 192.168.252.42 (extension). AskoziaPBX =>Accaunts => Phones, : Extension: 350 Caller ID: Lisa Password: 1234 Providers: 300user IAX2 PBX2 (192.168.252.41) PBX3 (192.168.252.42) .
∨ xx 1 x2 xx 4 (1* xx 3 *1 ∨1*0 ∨ 1*1) ∨ xx 1 x2 x4 (0* xx 3 *0 ∨ 1*0 ∨1*0) ∨ ∨ x1 xx 2 xx 4 (1* xx 3*1 ∨ 0*1 ∨ 0*1) ∨x1 xx 2 x4 (0* xx 3*0 ∨ 0*1 ∨ 0*0) ∨ ∨ x1 x2 xx 4 (1* xx 3 *1 ∨ 0*0 ∨ 0*1) ∨x1 x2 x4 (1* xx 3*0 ∨ 0*0 ∨ 0*0) = = xx 1 xx 2 xx 4 (0) ∨xx 1 xx 2 x4 (1) ∨ xx 1 x2 xx 4 ( 1) ∨ xx 1 x2 x4 (0) ∨ x1 xx 2 xx 4 (xx 3) ∨ ∨ x1 xx 2 x4 (0) ∨ x1 x2 xx 4 ( xx 3 ) ∨x1 x2 x4 (0)= = xx 1 xx 2 x4 ∨ xx 1 x2 xx 4 ∨ x1 xx 2 xx 3xx 4 ∨x1 x2 xx 3xx 4 8. Teha punktis 3 saadud MDNK-le Shannoni disjunktiivne arendus vabaltvalitud 2he muutuja järgi. Valin muutujateks x1 ja x2. x2 xx 3 xx 4 ∨ xx 1 xx 2 ∨ xx 1 xx 4 = = xx 1 xx 2 (0* xx 3 xx 4 ∨ 1*1∨ 1* xx 4) ∨ x1 xx 2 (0*xx 3 xx 4 ∨ 0*1 ∨ 0*xx 4) ∨ ∨xx 1 x2 (1* xx 3 xx 4 ∨ 1*0 ∨ 1*xx 4) ∨ x1 x2 (1*xx 3 xx 4 ∨ 0*0 ∨ 0* xx 4) = = xx 1 xx 2 (1) ∨ x1 xx 2 (0) ∨ xx 1 x2 ( xx 3 xx 4 ∨ xx 4) ∨ x1 x2 (xx 3 xx 4 ) =
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