1. Nested Quantifiers Goals: Explain how to work with nested quantifiers
Show that the order of quantification matters.
Work with logical expressions involving multiple quantifiers.

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Nested Iteration
Let the domain be { 1, 2, …, 10 }.
Let P( x, y ) denote x > y.
x y P( x, y ) means x ( y P( x, y ) )
Is the above statement true?
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3. Copyright © Peter Cappello
boolean axEyP() // x y P( x, y ) {
for ( int x = 1; x <= 10; x++ )
boolean b = false;
for ( int y = 1; y <= 10; y++ )
if ( x > y )
b = true;
break; // finding 1 y value is enough }
if ( ! b )
return false;
return true;
Computational
Interpretation
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Multiple Quantifiers
Legend: A
B is valid
x  y P(x, y)
y  x P(x, y)
y x P(x, y)
x y P(x, y)
x y P(x, y)
y x P(x, y)
y x P(x, y)
 x y P(x, y)
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Translate to English
Let the domain be the real numbers.
x y ( ( x ≥ 0  y < 0 )  x – y > 0 )
Is there something wrong with
x ( ( x ≥ 0  y ( y < 0 ) )  x – y > 0 )
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6. Translate to a Logical Expression
Let Q( s, q ) denote “s has been a contestant on quiz show q”
I( s1, s2 ) denote “student s1 is student s2”
The domain for s, s1, s2 is students at UCSB.
The domain for q is quiz shows on TV.
Give a logical expression for:
Every TV quiz show has had a student from UCSB as a contestant.
At least 2 students from UCSB have been contestants on Jeopardy.
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Translations
1. q s Q( s, q )
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8. Copyright © Peter Cappello
2. s1 s2 ( I( s1, s2 ) 
Q( s1, Jeopardy )  Q( s2 , Jeopardy ) )
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9. Negating Nested Quantifiers
Negate x y ( P( x, y )  Q( x, y ) ) so that no quantifiers are negated.
x y ( P( x, y )  Q( x, y ) ).
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10. Negating Nested Quantifiers
Negate x y ( P( x, y )  Q( x, y ) ) so that no quantifiers are negated.
x y ( P( x, y )  Q( x, y ) ).
x y ( P( x, y )  Q( x, y ) ).
Copyright © Peter Cappello

11. Negating Nested Quantifiers
Negate x y ( P( x, y )  Q( x, y ) ) so that no quantifiers are negated.
x y ( P( x, y )  Q( x, y ) ).
x y ( P( x, y )  Q( x, y ) ).
x y  ( P( x, y )  Q( x, y ) ).
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12. Negating Nested Quantifiers
Negate x y ( P( x, y )  Q( x, y ) ) so that no quantifiers are negated.
x y ( P( x, y )  Q( x, y ) ).
x y ( P( x, y )  Q( x, y ) ).
x y  ( P( x, y )  Q( x, y ) ).
x y (  P( x, y )   Q( x, y ) ).
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