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User:Jon Awbrey/SCRATCHPAD (view source)
Revision as of 14:34, 31 January 2009
, 14:34, 31 January 2009→Definition 5
===Definition 5===
===Definition 5===
====Variant 1====
<br>
then the following are identical propositions:
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|- style="height:40px; text-align:center"
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| width="20%" | <math>\text{Definition 5}\!</math>
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|- style="height:40px"
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| width="18%" style="border-top:1px solid black" | <math>\text{If}\!</math>
| width="80%" style="border-top:1px solid black" | <math>Q ~\subseteq~ X</math>
|- style="height:40px"
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| <math>\text{then}\!</math>
| <math>\text{the following are identical propositions:}\!</math>
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|- style="height:40px"
| width="2%" style="border-top:1px solid black" |
| width="18%" style="border-top:1px solid black" | <math>\operatorname{D5a.}</math>
| width="80%" style="border-top:1px solid black" | <math>\upharpoonleft Q \upharpoonright</math>
|- style="height:60px"
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| <math>\operatorname{D5b.}</math>
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<math>\begin{array}{lcl}
f & : & X \to \underline\mathbb{B}
\\
f(x) & = & \downharpoonleft x \in Q \downharpoonright \quad (\forall x \in X)
\end{array}</math>
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<br>
====Variant 2====
<br>
<br>
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|- style="height:60px"
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| <math>\operatorname{D5b.}</math>
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<math>\begin{array}{lcl}
<math>\begin{array}{ccccl}
f & : & X \to \underline\mathbb{B}
f & : & X & \to & \underline\mathbb{B}
\\
\\
f(x) & = & \downharpoonleft x \in Q \downharpoonright \quad (\forall x \in X)
f & : & x & \mapsto & \downharpoonleft x \in Q \downharpoonright
\end{array}</math>
\end{array}</math>
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