Key for Homework 2 on Selection

 Q  A   Reason
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 1  9   To print "one", (A && B) must be true, so A must be true and B must be true,
        and (!C || !D) must be true, so that C must be false or D must be false, or
        both must be false.
        
 2  8   To print "four", (A && B) must be false, so A must be false or B must be false,
        or both must be false, and (C != D) must be true, so C and D cannot both be 
        false.
        
 3  1   This is just tedious; check all the possible combinations of truth values
        for A, B, C and D.
        
 4 10   To print "three", (A && B) must be true, so A and B must both be true.  And
        (!C || !D) must be false [or "one" would be printed], so C and D must both
        be true.  But then the condition for printing "two" would be true and so
        that would be printed.  In fact, there's NO way for "three" to be printed.

 5  3   To print "Two", the first if condition must be false and the second if
        condition must be true.  That means that x >0 must be true, and also that
        x <= 10 must be true.

 6  6   If Enter were 4, then case 4 would be executed.  Since that's just a break,
        the value of Enter wouldn't be changed, so it would still be 4.
        
 7  2   If Enter were 1, then case 1 would executed, changing Enter to -4.  But
        there's no break in case 1, so execution would fall through to case 2, and
        Enter would be changed to -6.  Again, there's no break, so execution would
        fall through to case 4, and the break there would terminate the switch.
        
 8  3   Apply DeMorgan's Law:   !(x < 10 || x >= 14) == !(x < 10) && !(x >= 14)
                                                     ==   x >= 10 && x < 14

 9  5   Apply DeMorgan's Law:  !(A && !B || C) == !( (A && !B) || C )
                                               == !(A && !B) && !C
                                               == (!A || !!B) && !C
                                               == (!A || B) && !C
                                               
        Note:  the parentheses DO matter because && has higher precedence than ||.
        The closest given answer was 2, but it omits the parentheses.

10  4   Short-circuiting means that the Boolean expression in Line 1 will be
        be evaluated correctly, since the first clause will be false.  In turn,
        that means the second if-condition (Line 3) will also evaluate to
        false, and so the output statement in line 6 will be executed.