# Z dx for some integers c and d thus for any integers

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z = dx for some integers c and d . Thus, for any integers a and b , ay + bz = acx + bdx = ( ac + bd ) x . Since we know the product of integers is also an integer, ac and bd are both integers. By the closure property of integers under addition, the ac + bd must also be an integer. Thus, since ac + bd is an integer, we know that x | ay + bz Problem 3 (5 points) For ease of notation, we define sets A = { 1 , . . . , 9 } , B = { 0 , 1 , 2 } and C = { 1 , 4 , 7 } . R = x A. y A. z A. ( y 6 = z ) [ f ( x, y ) 6 = f ( x, z )] C = x A. y A. z A. ( y 6 = z ) [ f ( y, x ) 6 = f ( z, x )] S = x C. y C. r 1 B. c 1 B. r 2 B. c 2 B. [( r 1 6 = r 2 ) ( c 1 6 = c 2 )] [ f ( x + r 1 , y + c 1 ) 6 = f ( x + r 2 , y + c 2 )] R C S

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Problem 4 (12 points) 1. T ( Alice ) T ( Bob ) T ( Charlie ) 2. x.T ( x ) S ( x ) F ( x ) 3. ¬ [ x.F ( x ) L ( x, get _ up _ early )] 4. x.S ( x ) L ( x, sunny _ weather ) 5. y. [ L ( Charlie, y ) → ¬ L ( Alice, y )] [ ¬ L ( Charlie, y ) L ( Alice, y )] 6. L ( Charlie, get _ up _ early ) L ( Charlie, sunny _ weather ) 7. x.T ( x ) F ( x ) ∧ ¬ S ( x ) Proof : From 6, we know that Charlie likes sunny weather. Using this and 5, we know that Alice does not like sunny weather. Using this and 4, we know that Alice is not a surfer. Since Alice is a CSE student (statement 1), by statement 2 we know alice must be a surfer or a frisbee player. Since she is not a surfer, she is a frisbee player. Thus Alice is a CSE20 student and is a frisbee player but is not a surfer. Thus there exists a CSE20 student who is a frisbee player but not a surfer.
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