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MyWikiBiz, Author Your Legacy — Thursday November 14, 2024
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12 bytes removed ,  18:08, 11 December 2008
reorganize
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But the latter is not a theorem in anyone's philosophy, so there is really no disagreement here.
 
But the latter is not a theorem in anyone's philosophy, so there is really no disagreement here.
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===Functional quantifiers===
+
==Functional Quantifiers==
    
The '''relative umpire operator''' <math>\Upsilon : (\mathbb{B}^2 \to \mathbb{B})^2 \to \mathbb{B}</math> takes two propositions as arguments and gives the value <math>1\!</math> if and only if the first implies the second.  In symbols:
 
The '''relative umpire operator''' <math>\Upsilon : (\mathbb{B}^2 \to \mathbb{B})^2 \to \mathbb{B}</math> takes two propositions as arguments and gives the value <math>1\!</math> if and only if the first implies the second.  In symbols:
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<br>
 
<br>
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====Tables====
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===Tables===
    
The auxiliary notations:
 
The auxiliary notations:
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|}<br>
 
|}<br>
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====Exercises====
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===Exercises===
    
Express the following formulas in functional terms.
 
Express the following formulas in functional terms.
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=====Exercise 1=====
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====Exercise 1====
    
<blockquote>
 
<blockquote>
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Need to think a little more about the proposition <math>p \Rightarrow q</math> as a boolean function of type <math>\mathbb{B}^2 \to \mathbb{B}</math> and the corresponding higher order proposition of type <math>(\mathbb{B}^2 \to \mathbb{B}) \to \mathbb{B}.</math>
 
Need to think a little more about the proposition <math>p \Rightarrow q</math> as a boolean function of type <math>\mathbb{B}^2 \to \mathbb{B}</math> and the corresponding higher order proposition of type <math>(\mathbb{B}^2 \to \mathbb{B}) \to \mathbb{B}.</math>
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=====Exercise 2=====
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====Exercise 2====
    
<blockquote>
 
<blockquote>
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</blockquote>
 
</blockquote>
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=====Exercise 3=====
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====Exercise 3====
    
<blockquote>
 
<blockquote>
 
<math>(\forall x \in X)(Px \Rightarrow Qx) \lor (\forall x \in X)(Qx \Rightarrow Px)</math>
 
<math>(\forall x \in X)(Px \Rightarrow Qx) \lor (\forall x \in X)(Qx \Rightarrow Px)</math>
 
</blockquote>
 
</blockquote>
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