MyWikiBiz, Author Your Legacy — Wednesday April 09, 2025
Jump to navigationJump to search
12 bytes removed
, 18:08, 11 December 2008
Line 427:
Line 427:
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.
−
===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:
Line 437:
Line 437:
<br>
<br>
−
====Tables====
+
===Tables===
The auxiliary notations:
The auxiliary notations:
Line 758:
Line 758:
|}<br>
|}<br>
−
====Exercises====
+
===Exercises===
Express the following formulas in functional terms.
Express the following formulas in functional terms.
−
=====Exercise 1=====
+
====Exercise 1====
<blockquote>
<blockquote>
Line 778:
Line 778:
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>
−
=====Exercise 2=====
+
====Exercise 2====
<blockquote>
<blockquote>
Line 784:
Line 784:
</blockquote>
</blockquote>
−
=====Exercise 3=====
+
====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>