Changes

The same connection <math>F : \underline\mathbb{B}^2 \to \underline\mathbb{B}</math> can also be read as a proposition about things in the universe <math>X = \underline\mathbb{B}^2.</math>  If <math>s\!</math> is a sentence that denotes the proposition <math>F,\!</math> then the corresponding assertion says exactly what one states in uttering the sentence <math>^{\backprime\backprime} \, x ~\operatorname{is~not~equal~to}~ y \, ^{\prime\prime}.</math>  In such a case, one has <math>\downharpoonleft s \downharpoonright \, = F,</math> and all of the following expressions are ordinarily taken as equivalent descriptions of the same set:

The same connection <math>F : \underline\mathbb{B}^2 \to \underline\mathbb{B}</math> can also be read as a proposition about things in the universe <math>X = \underline\mathbb{B}^2.</math>  If <math>s\!</math> is a sentence that denotes the proposition <math>F,\!</math> then the corresponding assertion says exactly what one states in uttering the sentence <math>^{\backprime\backprime} \, x ~\operatorname{is~not~equal~to}~ y \, ^{\prime\prime}.</math>  In such a case, one has <math>\downharpoonleft s \downharpoonright \, = F,</math> and all of the following expressions are ordinarily taken as equivalent descriptions of the same set:

{| align="center" cellpadding="4" style="text-align:left" width="90%"

{| align="center" cellpadding="8" width="90%"

| &nbsp;

|

<math>\begin{array}{lll}

| <math>[| \downharpoonleft s \downharpoonright |]</math>

[| \downharpoonleft s \downharpoonright |]

| <math>=\!</math>

& = & [| F |]

| <math>[| F |]\!</math>

\\[6pt]

& = & F^{-1} (\underline{1})

| &nbsp;

| <math>=\!</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ s ~\}

| <math>F^{-1} (\underline{1})</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ F(x, y) = \underline{1} ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ F(x, y) ~\}

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ s ~\}</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \underline{(}~x~,~y~\underline{)} = \underline{1} ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \underline{(}~x~,~y~\underline{)} ~\}

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ F(x, y) = \underline{1} ~\}</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x ~\operatorname{exclusive~or}~ y ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \operatorname{just~one~true~of}~ x, y ~\}

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ F(x, y) ~\}</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x ~\operatorname{not~equal~to}~ y ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \underline{(}~x~,~y~\underline{)} = \underline{1} ~\}</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x \neq y ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x + y ~\}.

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \underline{(}~x~,~y~\underline{)} ~\}</math>

\end{array}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x ~\operatorname{exclusive~or}~ y ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ \operatorname{just~one~true~of}~ x, y ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x ~\operatorname{not~equal~to}~ y ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x \nLeftrightarrow y ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x \neq y ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ (x, y) \in \underline\mathbb{B}^2 ~:~ x + y ~\}.</math>

|}

|}

For each choice of propositions <math>p\!</math> and <math>q\!</math> about things in <math>X,\!</math> the stretch of <math>F\!</math> to <math>p\!</math> and <math>q\!</math> on <math>X\!</math> is just another proposition about things in <math>X,\!</math> a simple proposition in its own right, no matter how complex its current expression or its present construction as <math>F^\$ (p, q) = \underline{(}~p~,~q~\underline{)}^\$</math> makes it appear in relation to <math>p\!</math> and <math>q.\!</math>  Like any other proposition about things in <math>X,\!</math> it indicates a subset of <math>X,\!</math> namely, the fiber that is variously described in the following ways:

For each choice of propositions <math>p\!</math> and <math>q\!</math> about things in <math>X,\!</math> the stretch of <math>F\!</math> to <math>p\!</math> and <math>q\!</math> on <math>X\!</math> is just another proposition about things in <math>X,\!</math> a simple proposition in its own right, no matter how complex its current expression or its present construction as <math>F^\$ (p, q) = \underline{(}~p~,~q~\underline{)}^\$</math> makes it appear in relation to <math>p\!</math> and <math>q.\!</math>  Like any other proposition about things in <math>X,\!</math> it indicates a subset of <math>X,\!</math> namely, the fiber that is variously described in the following ways:

{| align="center" cellpadding="4" style="text-align:left" width="90%"

{| align="center" cellpadding="8" width="90%"

| &nbsp;

|

<math>\begin{array}{lll}

| <math>[| F^\$ (p, q) |]</math>

[| F^\$ (p, q) |]

| <math>=\!</math>

& = & [| \underline{(}~p~,~q~\underline{)}^\$ |]

| <math>[| \underline{(}~p~,~q~\underline{)}^\$ |]</math>

& = & (F^\$ (p, q))^{-1} (\underline{1})

| &nbsp;

| <math>=\!</math>

& = & \{~ x \in X ~:~ F^\$ (p, q)(x) ~\}

| <math>(F^\$ (p, q))^{-1} (\underline{1})</math>

& = & \{~ x \in X ~:~ \underline{(}~p~,~q~\underline{)}^\$ (x) ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ x \in X ~:~ \underline{(}~p(x)~,~q(x)~\underline{)} ~\}

| <math>\{~ x \in X ~:~ F^\$ (p, q)(x) ~\}</math>

& = & \{~ x \in X ~:~ p(x) + q(x) ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ x \in X ~:~ p(x) \neq q(x) ~\}

| <math>\{~ x \in X ~:~ \underline{(}~p~,~q~\underline{)}^\$ (x) ~\}</math>

& = & \{~ x \in X ~:~ \upharpoonleft P \upharpoonright (x) ~\neq~ \upharpoonleft Q \upharpoonright (x) ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ x \in X ~:~ x \in P ~\nLeftrightarrow~ x \in Q ~\}

| <math>\{~ x \in X ~:~ \underline{(}~p(x)~,~q(x)~\underline{)} ~\}</math>

& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\operatorname{or}~ x \in Q\!-\!P ~\}

| &nbsp;

| <math>=\!</math>

& = & \{~ x \in X ~:~ x \in P\!-\!Q ~\cup~ Q\!-\!P ~\}

| <math>\{~ x \in X ~:~ p(x) + q(x) ~\}</math>

& = & \{~ x \in X ~:~ x \in P + Q ~\}

| &nbsp;

| <math>=\!</math>

& = & P + Q ~\subseteq~ X

| <math>\{~ x \in X ~:~ p(x) \neq q(x) ~\}</math>

& = & [|p|] + [|q|] ~\subseteq~ X

| &nbsp;

\end{array}</math>

| <math>=\!</math>

| <math>\{~ x \in X ~:~ \upharpoonleft P \upharpoonright (x) ~\neq~ \upharpoonleft Q \upharpoonright (x) ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ x \in X ~:~ x \in P ~\nLeftrightarrow~ x \in Q ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ x \in X ~:~ x \in P\!-\!Q ~\operatorname{or}~ x \in Q\!-\!P ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ x \in X ~:~ x \in P\!-\!Q ~\cup~ Q\!-\!P ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>\{~ x \in X ~:~ x \in P + Q ~\}</math>

| &nbsp;

| <math>=\!</math>

| <math>P + Q ~\subseteq~ X</math>

| &nbsp;

| <math>=\!</math>

| <math>[|p|] + [|q|] ~\subseteq~ X</math>

|}

|}