- #1

- 315

- 34

**Homework Statement::**x

**Relevant Equations::**x

I stumbled upon the following example in the book - " How to prove it, A structured approach " ( 2nd edition) , Vellerman.

**Homework Statement::**

He then asks to describe the set:

## \bigcup_{s \in S} L_{s} \, \backslash \, \bigcup_{s \in S} A_{s} ##

**Relevant Equations::**

## L_{s} = \{ t \in S | \, L(s,t) \} ##

## A_{s} = \{ t \in S | \, A(s,t) \} ##

and It was also defined beforehand in the book that:

## \bigcup_{s \in S} L_{s} = \{t\,| \exists s \in S \, ( t \in L_{s} ) \} \, = \, \{t \in S | \, \exists s \in S \, L(s,t) \} ##

## \bigcup_{s \in S} A_{s} = \{t\,| \exists s \in S \, ( t \in A_{s} ) \} \, = \, \{t \in S | \, \exists s \in S \, A(s,t) \} ##

**Vellerman's answer to the question:**

## \bigcup_{s \in S} L_{s} \, \backslash \, \bigcup_{s \in S} A_{s} = \{t\,| t \in \bigcup_{s \in S} L_{s} \, \land \, t \notin \bigcup_{s \in S} A_{s} \} \, ##

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**Attempt at a solution::**

I have arrived to the same answer, however, I have decided to venture further and write the statement logically in its most concise form,

so the following is my answer, I would highly appreciate if you could tell me if I was correct in my reasoning and my answer as written below:

Given:

## L_{s} = \{ t \in S | \, L(s,t) \} ##

## A_{s} = \{ t \in S | \, A(s,t) \} ##

It is also known that:

## \bigcup_{s \in S} L_{s} = \{t\,| \exists s \in S \, ( t \in L_{s} ) \} \, = \, \{t \in S | \, \exists s \in S \, L(s,t) \} ##

## \bigcup_{s \in S} A_{s} = \{t\,| \exists s \in S \, ( t \in A_{s} ) \} \, = \, \{t \in S | \, \exists s \in S \, A(s,t) \} ##

I write the last statements as:

## \bigcup_{s \in S} L_{s} = \{t\,| ( \exists s \in S \, L(s,t) ) \land ( t \in S ) \} \, ##

## \bigcup_{s \in S} A_{s} = \{t\,| ( \exists s \in S \, A(s,t) ) \land ( t \in S ) \} \, ##

Thus

## \bigcup_{s \in S} L_{s} \, \backslash \, \bigcup_{s \in S} A_{s} = \{t\,| t \in \bigcup_{s \in S} L_{s} \, \land \, t \notin \bigcup_{s \in S} A_{s} \} \, ## = ## \forall t ( t \in \bigcup_{s \in S} L_{s} \, \land \, t \notin \bigcup_{s \in S} A_{s} ) ##

## = \forall t ( [ ( \exists s \in S \, L(s,t) ) \land ( t \in S ) ] \land \neg [( \exists s \in S \, A(s,t) ) \land ( t \in S ) ] ) ##

(changing the bound variable from s to b in the second half of the expression to avoid confusion, thus: ) ## = \forall t ( [ ( \exists s \in S \, L(s,t) ) \land ( t \in S ) ] \land \neg [( \exists b \in S \, A(b,t) ) \land ( t \in S ) ] ) ##

( using the identity ## \neg ( P \land Q ) = \neg P \lor \neg Q ## , thus: )

## = \forall t ( [ ( \exists s \in S \, L(s,t) ) \land ( t \in S ) ] \land [\neg( \exists b \in S \, A(b,t) ) \lor ( t \notin S ) ] ) ##

( using the distributivity property, thus: )

## = \forall t ( [ ( \exists s \in S \, L(s,t) ) \land ( t \in S ) \land \neg( \exists b \in S \, A(b,t) ) ] \lor [ ( \exists s \in S \, L(s,t) ) \land ( t \in S ) \land ( t \notin S ) ] ) ##

## = \forall t ( ( \exists s \in S \, L(s,t) ) \land ( t \in S ) \land \neg( \exists b \in S \, A(b,t) ) ) ##

In english, this means: the set of all students who are liked by at least one student, but are not admired by any students.

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Was I correct in writing the final logical form of the answer? ( I can't verify this as I'm self-studying )