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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)

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)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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For numbers 6 – 7. Given: If p, q, r, and s denote the following propositions. p: 1+1#3 q: 6. " If Alexander exists or x2 - 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 r: Alexander exists x2 – 2x + 1 = 0, where x = 1 s: Mouse is an output device (rVq) → - (~ p^ ~s) (rVq) → - (-p as) C. (rVq) → -( pns) A. В. D. (rVq) → (-p^~s) 7. " Either mouse is an output device and x? - 2x + 1 # 0, where x = 1, or 1+ 1 = 3 and Alexander does not exists" can be symbolized as (s^q) V (~p^ ~r) В. C. (s^~q) V (p ^~r) A. (s a q) V (p ^~r) D. (S ^-q) V (~p ^~r)
Transcribed Image Text:For numbers 6 – 7. Given: If p, q, r, and s denote the following propositions. p: 1+1#3 q: 6. " If Alexander exists or x2 - 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 r: Alexander exists x2 – 2x + 1 = 0, where x = 1 s: Mouse is an output device (rVq) → - (~ p^ ~s) (rVq) → - (-p as) C. (rVq) → -( pns) A. В. D. (rVq) → (-p^~s) 7. " Either mouse is an output device and x? - 2x + 1 # 0, where x = 1, or 1+ 1 = 3 and Alexander does not exists" can be symbolized as (s^q) V (~p^ ~r) В. C. (s^~q) V (p ^~r) A. (s a q) V (p ^~r) D. (S ^-q) V (~p ^~r)
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