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→‎Derived Equivalence Relations: more white space in ordered list
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The relation <math>\operatorname{Der}(L)</math> is defined and the notation <math>x ~\overset{L}{=}~ y</math> is meaningful in every situation where the corresponding denotation operator <math>\operatorname{Den}(-,-)</math> makes sense, but it remains to check whether this relation enjoys the properties of an equivalence relation.
 
The relation <math>\operatorname{Der}(L)</math> is defined and the notation <math>x ~\overset{L}{=}~ y</math> is meaningful in every situation where the corresponding denotation operator <math>\operatorname{Den}(-,-)</math> makes sense, but it remains to check whether this relation enjoys the properties of an equivalence relation.
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# Reflexive property. Is it true that <math>x ~\overset{L}{=}~ x</math> for every <math>x \in S = I</math>? By definition, <math>x ~\overset{L}{=}~ x</math> if and only if <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, x).</math> Thus, the reflexive property holds in any setting where the denotations <math>\operatorname{Den}(L, x)</math> are defined for all signs <math>x\!</math> in the syntactic domain of <math>R.\!</math>
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<ol style="list-style-type:decimal">
# Symmetric property. Does <math>x ~\overset{L}{=}~ y ~\Rightarrow~ y ~\overset{L}{=}~ x</math> for all <math>x, y \in S</math>? In effect, does <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, y)</math> imply <math>\operatorname{Den}(L, y) = \operatorname{Den}(L, x)</math> for all signs <math>x\!</math> and <math>y\!</math> in the syntactic domain <math>S\!</math>? Yes, so long as the sets <math>\operatorname{Den}(L, x)</math> and <math>\operatorname{Den}(L, y)</math> are well-defined, a fact which is already being assumed.
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# Transitive property. Does <math>x ~\overset{L}{=}~ y</math> and <math>y ~\overset{L}{=}~ z</math> imply <math>x ~\overset{L}{=}~ z</math> for all <math>x, y, z \in S</math>? To belabor the point, does <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, y)</math> and <math>\operatorname{Den}(L, y) = \operatorname{Den}(L, z)</math> imply <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, z)</math> for all <math>x, y, z \in S</math>? Yes, again, under the stated conditions.
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<li>
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<p>'''Reflexive property.'''</p>
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<p>Is it true that <math>x ~\overset{L}{=}~ x</math> for every <math>x \in S = I</math>?</p>
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<p> By definition, <math>x ~\overset{L}{=}~ x</math> if and only if <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, x).</math></p>
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<p>Thus, the reflexive property holds in any setting where the denotations <math>\operatorname{Den}(L, x)</math> are defined for all signs <math>x\!</math> in the syntactic domain of <math>R.\!</math></p></li>
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<li>
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<p>'''Symmetric property.'''</p>
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<p>Does <math>x ~\overset{L}{=}~ y ~\Rightarrow~ y ~\overset{L}{=}~ x</math> for all <math>x, y \in S</math>?</p>
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<p>In effect, does <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, y)</math> imply <math>\operatorname{Den}(L, y) = \operatorname{Den}(L, x)</math> for all signs <math>x\!</math> and <math>y\!</math> in the syntactic domain <math>S\!</math>?</p>
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<p>Yes, so long as the sets <math>\operatorname{Den}(L, x)</math> and <math>\operatorname{Den}(L, y)</math> are well-defined, a fact which is already being assumed.</p></li>
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<li>
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<p>'''Transitive property.'''</p>
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<p>Does <math>x ~\overset{L}{=}~ y</math> and <math>y ~\overset{L}{=}~ z</math> imply <math>x ~\overset{L}{=}~ z</math> for all <math>x, y, z \in S</math>?</p>
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<p>To belabor the point, does <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, y)</math> and <math>\operatorname{Den}(L, y) = \operatorname{Den}(L, z)</math> imply <math>\operatorname{Den}(L, x) = \operatorname{Den}(L, z)</math> for all <math>x, y, z \in S</math>?</p>
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<p>Yes, once again, under the stated conditions.</p></li>
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</ol>
    
It should be clear at this point that any question about the equiference of signs reduces to a question about the equality of sets, specifically, the sets that are indexed by these signs.  As a result, so long as these sets are well-defined, the issue of whether equiference relations induce equivalence relations on their syntactic domains is almost as trivial as it initially appears.
 
It should be clear at this point that any question about the equiference of signs reduces to a question about the equality of sets, specifically, the sets that are indexed by these signs.  As a result, so long as these sets are well-defined, the issue of whether equiference relations induce equivalence relations on their syntactic domains is almost as trivial as it initially appears.
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