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69 | 69 | only the !!{element}s of $\Top{O}$ are. $!A \Entails !B$ iff $!B$ is
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70 | 70 | true at every point at which~$!A$ is true, i.e., $\Prop{X}{!A}
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71 | 71 | \subseteq \Prop{X}{!B}$, for all~$X$. The absurd statement~$\lfalse$
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72 |
| -is never true, so $\Prop{X}{\lfalse} = \emptyset$. How must the |
73 |
| -propositions expressed by $!B \land !C$, $!B \lor !C$, and $!B \lif |
74 |
| -!C$ be related to those expressed by $!B$ and~$!C$ for the |
75 |
| -intuitionistically valid laws to hold, i.e., so that $!A \Proves !B$ |
76 |
| -iff $\Prop{X}{!A} \subset \Prop{X}{!B}$. $\lfalse \Proves !A$ for any |
77 |
| -$!A$, and only $\emptyset \subseteq U$ for all $U$. Since $!B \land |
78 |
| -!C \Proves !B$, $\Prop{X}{!B \land !C} \subseteq \Prop{X}{!B}$, and |
79 |
| -similarly $\Prop{X}{!B \land !C} \subseteq \Prop{X}{!C}$. The largest |
80 |
| -set satisfying $W \subseteq U$ and $W \subseteq V$ is $U \cap V$. |
| 72 | +is never true, so $\Prop{X}{\lfalse} = \emptyset$. |
| 73 | + |
| 74 | +How must the propositions expressed by $!B \land !C$, $!B \lor !C$, |
| 75 | +and $!B \lif !C$ be related to those expressed by $!B$ and~$!C$ for |
| 76 | +the intuitionistically valid laws to hold, i.e., so that $!A \Proves |
| 77 | +!B$ iff $\Prop{X}{!A} \subset \Prop{X}{!B}$? We require $\lfalse |
| 78 | +\Proves !A$ for any $!A$, which is satisfied because $\emptyset |
| 79 | +\subseteq U$ for all $U$. Since $!B \land !C \Proves !B$, we require |
| 80 | +that $\Prop{X}{!B \land !C} \subseteq \Prop{X}{!B}$, and similarly |
| 81 | +$\Prop{X}{!B \land !C} \subseteq \Prop{X}{!C}$. The largest set |
| 82 | +satisfying $W \subseteq U$ and $W \subseteq V$ is $U \cap V$. |
81 | 83 | Conversely, $!B \Proves !B \lor !C$ and $!C \Proves !B \lor !C$, and
|
82 |
| -so $\Prop{X}{!B} \subseteq \Prop{X}{!B \lor !C}$ and $\Prop{X}{!C} |
83 |
| -\subseteq \Prop{X}{!B \lor !C}$. The smallest set~$W$ such that $U |
84 |
| -\subseteq W$ and $V \subseteq W$ is $U \cup V$. The definition for |
85 |
| -$\lif$ is tricky: $!A \lif !B$ expresses the weakest proposition that, |
86 |
| -combined with $!A$, entails $!B$. That $!A \lif !B$ combined with $!A$ |
87 |
| -entails~$!B$ is clear from $(!A \lif !B) \land !A \Proves !B $. So |
88 |
| -$\Prop{X}{!A \lif !B}$ should be the greatest open set such that |
89 |
| -$\Prop{X}{!A \lif !B} \cap \Prop{X}{!A} \subset \Prop{X}{!B}$, leading |
90 |
| -to our definition. |
| 84 | +so we require that $\Prop{X}{!B} \subseteq \Prop{X}{!B \lor !C}$ and |
| 85 | +$\Prop{X}{!C} \subseteq \Prop{X}{!B \lor !C}$. The smallest set~$W$ |
| 86 | +such that $U \subseteq W$ and $V \subseteq W$ is $U \cup V$. |
| 87 | + |
| 88 | +The definition for $\lif$ is tricky: $!A \lif !B$ expresses the |
| 89 | +weakest proposition that, combined with $!A$, entails $!B$. That $!A |
| 90 | +\lif !B$ combined with $!A$ entails~$!B$ is clear from $(!A \lif !B) |
| 91 | +\land !A \Proves !B $. So $\Prop{X}{!A \lif !B}$ should be the |
| 92 | +greatest open set such that $\Prop{X}{!A \lif !B} \cap \Prop{X}{!A} |
| 93 | +\subset \Prop{X}{!B}$, leading to our definition. |
91 | 94 |
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92 | 95 | \end{document}
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