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COS1501 Assignment 2 2021

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UNISA COS1501 Theoretical Computer Science Assignment TWO of 2021 solutions.

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  • June 18, 2021
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COS1501 Assignment 2 2021



Question 1

[(𝐴 ∩ 𝐵)′ − 𝐶] ∩ [(𝐴 + 𝐵) − 𝐶]




𝐴∩𝐵




(𝐴 ∩ 𝐵)′ (𝐴 + 𝐵)




(𝐴 ∩ 𝐵)′ − 𝐶 (𝐴 + 𝐵) − 𝐶

,[(𝐴 ∩ 𝐵)′ − 𝐶] ∩ [(𝐴 + 𝐵) − 𝐶]




Question 1 FOUR



Question 2



𝑈 = {1, 2, 3, 4}
𝐴, 𝐵, 𝐶 𝑎𝑟𝑒 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝑈


(𝐴 − 𝐵) ∪ 𝐶 ′ = (𝐶 ′ − 𝐵) + 𝐴



1. 𝐴 = {1}, 𝐵 = {2} 𝑎𝑛𝑑 𝐶 = {3}

𝐿𝐻𝑆 = (𝐴 − 𝐵) ∪ 𝐶 ′
𝐿𝐻𝑆 = (𝐴 − 𝐵) ∪ (𝑈 − 𝐶)
𝐿𝐻𝑆 = ({1} − {2}) ∪ ({1, 2, 3, 4} − {3})
𝐿𝐻𝑆 = {1} ∪ {1, 2, 4}
𝐿𝐻𝑆 = {1, 2, 4}

𝑅𝐻𝑆 = (𝐶 ′ − 𝐵) + 𝐴
𝑅𝐻𝑆 = ((𝑈 − 𝐶) − 𝐵) + 𝐴
𝑅𝐻𝑆 = (({1, 2, 3, 4} − {3}) − {2}) + {1}
𝑅𝐻𝑆 = ({1, 2, 4} − {2}) + {1}
𝑅𝐻𝑆 = {1, 4} + {1}
𝑅𝐻𝑆 = {4}

, 𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆

𝑇ℎ𝑖𝑠 𝑐ℎ𝑜𝑖𝑐𝑒 𝑜𝑓 𝐴, 𝐵 𝑎𝑛𝑑 𝐶 𝑐𝑎𝑛 𝑏𝑒 𝑢𝑠𝑒𝑑 𝑎𝑠 𝑎 𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑒𝑥𝑎𝑚𝑝𝑙𝑒 𝑡𝑜 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡
(𝐴 − 𝐵) ∪ 𝐶 ′ = (𝐶 ′ − 𝐵) + 𝐴 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑛 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦.


1 is TRUE.

2. 𝐴 = {1}, 𝐵 = {1} 𝑎𝑛𝑑 𝐶 = {2}

𝐿𝐻𝑆 = (𝐴 − 𝐵) ∪ 𝐶 ′
𝐿𝐻𝑆 = (𝐴 − 𝐵) ∪ (𝑈 − 𝐶)
𝐿𝐻𝑆 = ({1} − {1}) ∪ ({1, 2, 3, 4} − {2})
𝐿𝐻𝑆 = ∅ ∪ {1, 3, 4}
𝐿𝐻𝑆 = {1, 3, 4}

𝑅𝐻𝑆 = (𝐶 ′ − 𝐵) + 𝐴
𝑅𝐻𝑆 = ((𝑈 − 𝐶) − 𝐵) + 𝐴
𝑅𝐻𝑆 = (({1, 2, 3, 4} − {2}) − {1}) + {1}
𝑅𝐻𝑆 = ({1, 3, 4} − {1}) + {1}
𝑅𝐻𝑆 = {3, 4} + {1}
𝑅𝐻𝑆 = {1, 3, 4}

𝐿𝐻𝑆 = 𝑅𝐻𝑆

𝑇ℎ𝑖𝑠 𝑐ℎ𝑜𝑖𝑐𝑒 𝑜𝑓 𝐴, 𝐵 𝑎𝑛𝑑 𝐶 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑢𝑠𝑒𝑑 𝑎𝑠 𝑎 𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑒𝑥𝑎𝑚𝑝𝑙𝑒 𝑡𝑜 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡
(𝐴 − 𝐵) ∪ 𝐶 ′ = (𝐶 ′ − 𝐵) + 𝐴 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑛 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦.


2 is FALSE.


3. 𝐴 = {1, 2}, 𝐵 = {1, 2} 𝑎𝑛𝑑 𝐶 = {3}

𝐿𝐻𝑆 = (𝐴 − 𝐵) ∪ 𝐶 ′
𝐿𝐻𝑆 = (𝐴 − 𝐵) ∪ (𝑈 − 𝐶)
𝐿𝐻𝑆 = ({1, 2} − {1, 2}) ∪ ({1, 2, 3, 4} − {3})
𝐿𝐻𝑆 = ∅ ∪ {1, 2, 4}
𝐿𝐻𝑆 = {1, 2, 4}

𝑅𝐻𝑆 = (𝐶 ′ − 𝐵) + 𝐴
𝑅𝐻𝑆 = ((𝑈 − 𝐶) − 𝐵) + 𝐴
𝑅𝐻𝑆 = (({1, 2, 3, 4} − {3}) − {1, 2}) + {1, 2}
𝑅𝐻𝑆 = ({1, 2, 4} − {1, 2}) + {1, 2}
𝑅𝐻𝑆 = {4} + {1, 2}
𝑅𝐻𝑆 = {1, 2. 4}

𝐿𝐻𝑆 = 𝑅𝐻𝑆

𝑇ℎ𝑖𝑠 𝑐ℎ𝑜𝑖𝑐𝑒 𝑜𝑓 𝐴, 𝐵 𝑎𝑛𝑑 𝐶 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑢𝑠𝑒𝑑 𝑎𝑠 𝑎 𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑒𝑥𝑎𝑚𝑝𝑙𝑒 𝑡𝑜 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡
(𝐴 − 𝐵) ∪ 𝐶 ′ = (𝐶 ′ − 𝐵) + 𝐴 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑛 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦.

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