6. " If Alexander exists or x? – 2x + 1 = 0, where x = 1 then it is not true that 1+1 = 3 and mouse is not an output device, " can be symbolized as (rVq) → - (~ p^ ~s) irva) → -(-pas) A. C. (rVq) → - ( p as) В. D. (rvq) → (-p ^~s) 7. " Either mouse is an output device and x2 – 2x + 1 # 0, where x = 1, or 1+ 1 = 3 and Alexander does not exists" can be symbolized as (Saq) V (~p^ ~r) (s ^-q) V (p ^~r) A. C. В. (sa q) V (p ^~ r) D. (S ^~q) V (~p a~r)

= 2 Simplify -(s^(mv-s)) to ¬sv-m meeee Jump to level 1 1. Select a law from the right to apply -(SA(MV-s)) W S X Check # 3 с D C Next 123 $ 4 R F 1 G Search or type URL % 5 V T G A 6 ♫A B MacBook Pro Y H 87 & M Distributive (a^b)v(a^c) (avb)^(avc) = Commutative avb aлb De Morgan's -(a^b) (avb) U Conditional a→b a+b J * 8 = + ( Laws a^(bvc) av(b^c) ¬аvb 9 K bva ba -av-b -а^-b (a+b)^(b→a) 0 Complement av ¬a L а па -T -F Identity алт а - T = F - F avF - а Double negation =a = T -a P W

Prove with usual notations that

Simplify (-nvq)-(n^q) to -n 1. Select a law from the right to apply (-nvq)^-(n^q) Distributive (a^b)v(a^c) (avb)^(avc) Commutative avb аль De Morgan's (аль) (avb) Conditional a→b a+b ||| ||| ||| ||| Laws a^(bvc) av(bлc) bva bдa -av-b -ал-Б avb =(a+b)^(b-a) Complement av ¬a = T алла Е F ¬T E -F ET Identity алт 3 а avF = a Double negation ла = a