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Let $x \in A$ be logic statement $p$ and $x \in B$ be statement $q$. Prove that $(A \oplus B)'= A' \oplus B$ using the fact that LHS and RHS have identical truth tables.

For LHS: $$ (A \oplus B)' = (A' \cup B) \cap (B' \cup A)\iff \{x:(x\notin A \lor x \in B) \land (x \notin B \lor x \in A) \} \\\iff (\lnot p \lor q) \land (\lnot q \lor p) $$ While for RHS: $$ A' \oplus B = (A' \cap B') \cup (B \cap A) \iff \{x: (x \notin A \land x \notin B) \lor (x \in B \land x\in A)\} \\\iff (\lnot p \land \lnot q) \lor (p \land q) $$ Using Wolfram Alpha the logical statements have the same truth tables (LHS and RHS).

But as far as I know two sets are equal iff they have the same elements. How can I use the fact of identical truth tables to help me?

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    $\begingroup$ Your have a logical (or at least notational) flaw here: "one set equals some other set is logically equivalent to some set, is logically equivalent to a statement" make s no sense. $\endgroup$ – Hagen von Eitzen Mar 31 '18 at 10:04
  • $\begingroup$ @HagenvonEitzen I wanted to say that $(A \oplus B)'$ means that/is equivalent to the logical statement. I'm not sure what notation should be used for this. $\endgroup$ – Yos Mar 31 '18 at 10:06
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    $\begingroup$ The notation is that the set construction is the set, so for instance : $$A\cap B {~= \{x: x\in A\wedge x\in B\} \\~= \{x: p\wedge q\}}$$ That is: the union of sets A and B is the set of entities for which the statement $p\wedge q$ is satisfied. $\endgroup$ – Graham Kemp Mar 31 '18 at 10:27
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The principle is that $\{x: P(x)\} = \{x: Q(x)\}$ if and only if $\forall x:( P(x)\leftrightarrow Q(x))$.   That is: the sets will be equal exactly when the predicates that construct them are identical.

The predicates describe which elements are(or are not) in the set, after all.

You should have $$\begin{align}(A\oplus B)' &= \{x: \neg((x\in A \wedge x\notin B)\vee(x\notin A\wedge x\in B))\} &&\text{by definition of }\oplus \\ &= \{x: \neg ((p\wedge \neg q)\vee(\neg p\wedge q))\} \\ &= \{x: (\neg p\vee q)\wedge(p\vee\neg q))\} \\ &~~\vdots &&\text{see table.}\\ & = \{x: (\neg p\wedge \neg q)\vee(p\wedge q)\}\\ &= \{x: (x\in A'\wedge x\notin B)\vee(x\notin A'\wedge x\in B)\} \\ &= (A'\oplus B) &&\text{by definition of }\oplus \end{align}$$

$\bbox[lemonchiffon,border:1pt solid blue]{\begin{array}{c:c|c:c:c:c|c:c} p & q & p\wedge q & p\wedge \neg q & \neg p\wedge q & \neg p\wedge \neg q & \neg( (p\wedge\neg q)\vee (\neg p\wedge q)) & (\neg p\wedge \neg q)\vee (p\wedge q)\\ \hline \top & \top & \top & \bot & \bot & \bot & \top & \top\\ \hdashline \top & \bot \\ \hdashline \bot & \top \\ \hdashline \bot &\bot \end{array}}$

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  • $\begingroup$ If I complete the "why" step in your explanation then I wouldn't need truth tables. As far as I understand I should arrive to the step when $(A \oplus B)'=...= \{x : (\lnot p \lor q) \land (\lnot q \lor p)\}$ and then separately $(A' \oplus B) = ... = (\lnot p \land \lnot q) \lor (p \land q)$ and then I can say that because the truth tables are equal therefore the two sets are equal because as you mentioned $\{x: P(x)\} = \{x: Q(x)\}$ iff $\forall x:( P(x)\leftrightarrow Q(x))$ $\endgroup$ – Yos Mar 31 '18 at 10:38
  • $\begingroup$ You don't need truth tables, @Yos . Use distribution. $\endgroup$ – Graham Kemp Mar 31 '18 at 11:19
  • $\begingroup$ That's definitely one way to solve it but in the OP I'm explicitly asking for a way to solve it using truth tables. $\endgroup$ – Yos Mar 31 '18 at 11:19
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    $\begingroup$ @Yos Then just fill in a truth table and verify the relevant columns have the same value in every row. $\endgroup$ – Graham Kemp Mar 31 '18 at 11:31
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Show that $$x\in(A\oplus B)'\iff x\in A'\oplus B$$ by using the definitions of $\oplus$ and ${}'$ to convert both sides into logical combinations of $p$ and $q$.

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