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Q1. 1. 2. 3. 4. 5. 6. 7. Q2.

(â&#x2C6;&#x20AC;đ?&#x2018;Ľ)(~đ?&#x2018; đ?&#x2018;Ľ â&#x160;&#x192; đ?&#x2018;&#x192;đ?&#x2018;Ľ) (â&#x2C6;&#x20AC;đ?&#x2018;Ľ)(đ??ˇđ?&#x2018;Ľ1 & đ??ˇđ?&#x2018;Ľ2 & đ??ˇđ?&#x2018;Ľ3 & đ??ˇđ?&#x2018;Ľ4 â&#x160;&#x192; ~đ?&#x2018; đ?&#x2018;Ľ) (â&#x2C6;&#x20AC;đ?&#x2018;Ś)(đ??ľđ?&#x2018;Ľđ?&#x2018;Ś & (đ?&#x2018;&#x2030;đ?&#x2018;Ľ3 â&#x2039; đ?&#x2018;&#x2030;đ?&#x2018;Ľ4) [(â&#x2C6;&#x192;đ?&#x2018;Ľ)(đ??żđ?&#x2018;Ľ1 &đ?&#x2018;&#x2020;đ?&#x2018;Ľ3)] â&#x160;&#x192; đ?&#x2018;&#x2020;13 đ?&#x2018;¤â&#x201E;&#x17D;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ??żđ?&#x2018;&#x17D;đ?&#x2018;?="a>b" đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x2020;đ?&#x2018;&#x17D;đ?&#x2018;? ="a<b" (â&#x2C6;&#x20AC;đ?&#x2018;Ľ)đ?&#x2018;&#x192;đ?&#x2018;Ľ â&#x160;&#x192; (â&#x2C6;&#x192;đ?&#x2018;§)(đ?&#x2018; đ?&#x2018;§ & đ?&#x2018;&#x2030;đ?&#x2018;Ľđ?&#x2018;§) (â&#x2C6;&#x20AC;đ?&#x2018;Ľ)(â&#x2C6;&#x192;đ?&#x2018;Ś)[(đ??śđ?&#x2018;Ľđ?&#x2018;Ś â&#x2039; đ?&#x2018;&#x2026;đ?&#x2018;Ľđ?&#x2018;Ś) â&#x2039; ~(â&#x2C6;&#x192;đ?&#x2018;§)(đ??ľđ?&#x2018;Ľđ?&#x2018;§)] đ??śđ?&#x2018;Ľđ?&#x2018;Ś â&#x160;&#x192; đ??śđ?&#x2018;Śđ?&#x2018;Ľ

1. 1.2 is true, as the domain covers only the integers 1 to 4 and the grid positions. So any object in the language must be one of those things. 1.6 is also true, since every grid position must be in a column or a row or else it is not a grid position. If it is not a grid position, it cannot be in a block, and so not being in a block follows from not being in a column or row. 2. Reflexive Irreflexive Symmetric Transitive â&#x153;&#x201C; â&#x153;&#x201C; â&#x153;&#x201C; B â&#x153;&#x201C; â&#x153;&#x201C; â&#x153;&#x201C; R â&#x153;&#x201C; â&#x153;&#x201C; â&#x153;&#x201C; C â&#x153;&#x201C; â&#x153;&#x201C; D â&#x153;&#x201C; V Q3. 1. The statement could be interpreted as â&#x20AC;&#x153;For all natural numbers x and y, if x is bigger than y, then y is x+1,â&#x20AC;? which is false. A counterexample is x=1, y=0, where we have the antecedent (1>0) but not the consequent (0 is not 1+1), and so the conditional is false. 2. The statement could be interpreted as â&#x20AC;&#x153;For all natural numbers x and y, if y is x+1, then x is bigger than y,â&#x20AC;? which is false. A counterexample is y=1, x=0, where we have the antecedent (1=0+1) but not the consequent (0 is not bigger than 1), and so the conditional is false. 3. The statement could be interpreted as â&#x20AC;&#x153;All even numbers are one greater than an odd number,â&#x20AC;? which is true. Consider an even number N. N must be divisible by 2, that is, N/2 is a natural number. However, (N-1)/2 = N/2 - ½. N/2 is a natural number, but ½ is not, therefore (N-1)/2 is not a natural number, and so N-1 is not divisible by 2. Hence, N-1 is odd. (QED) 4. The statement could be interpreted as â&#x20AC;&#x153;There is an odd number smaller than any even number,â&#x20AC;? which is true. 2 is the smallest even number. 1 is an odd number. 1<2. Therefore 1 is smaller than any even number. (QED) Q4. 1. 2. 3. 4.

** (â&#x2C6;&#x20AC;đ?&#x2018;&#x17D;)(â&#x2C6;&#x20AC;đ?&#x2018;?)[đ?&#x2018;&#x192;đ?&#x2018;&#x17D; &đ?&#x2018;&#x192;đ?&#x2018;? &đ??ˇđ?&#x2018;&#x17D;đ?&#x2018;? & đ?&#x2018; đ?&#x2018;Ľ & đ?&#x2018;&#x2030;đ?&#x2018;&#x17D;đ?&#x2018;Ľ & đ?&#x2018;&#x2030;đ?&#x2018;?đ?&#x2018;Ľ â&#x160;&#x192; ~đ??ľđ?&#x2018;&#x17D;đ?&#x2018;?] (â&#x2C6;&#x20AC;đ?&#x2018;§)(â&#x2C6;&#x20AC;đ?&#x2018;¤)(â&#x2C6;&#x20AC;đ?&#x2018;Ľ)(â&#x2C6;&#x192;đ?&#x2018;Ś)[đ?&#x2018; đ?&#x2018;§ & đ?&#x2018; đ?&#x2018;¤ & đ?&#x2018;&#x2030;đ?&#x2018;Ľđ?&#x2018;§ & đ?&#x2018;&#x2030;đ?&#x2018;Śđ?&#x2018;¤ & đ??ˇđ?&#x2018;Śđ?&#x2018;Ľ & (đ??ľđ?&#x2018;Ľđ?&#x2018;Ś â&#x2039; đ??śđ?&#x2018;Ľđ?&#x2018;Ś â&#x2039; đ?&#x2018;&#x2026;đ?&#x2018;Ľđ?&#x2018;Ś) â&#x160;&#x192; đ??ˇđ?&#x2018;§đ?&#x2018;¤] **

Assign3  

Assignment 3

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