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COS1501 Assignment 2 2021
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- University Of South Africa (Unisa)
UNISA COS1501 Theoretical Computer Science Assignment TWO of 2021 solutions.
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COS1501 Assignment 2 2021Question 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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