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Construct another English form sentence, which is logically equivalent to that which was given.

"Susan goes to school or Susan does not talk on the phone or Susan does not go to school."

Mike Shasaco
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  • You might be interested in how to post logical/mathematical expressions using MathJax and $\LaTeX$, esp. if you decide to post your own attempt at a solution. – hardmath Jun 19 '16 at 19:04

2 Answers2

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Let $A$ be a statement "Susan goes to school" and $B$ be a stetement "Susan does not talk on the telephone". Then statement "Susan goes to school or Susan does not talk on the phone or Susan does not go to school" may be represented as $A\lor B\lor\neg A$ where $\lor$ is "or" and $\neg$ is "not". Now we have $$ A\lor B\lor\neg A \equiv A\lor\neg A\lor B \equiv 1\lor B \equiv 1, $$ where $1$ is truth. So your statement is equvivalent to logical truth, i.e. this statement is always true.

Indeed, regerdless of whether Susan talks on the phone or not, she definetly goes or not goes to school, so whole statement is true.

Why $x\lor \neg x \equiv 1$ and $1 \lor x \equiv 1$

One may also use the truth tale to show this. Let $1$ be a truth and $0$ be a false. Truth table of "or" operator $\vee$ is $$ \begin{array}{cc|c} x & y & x\lor y \\ \hline 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{array} $$ Here $x$ and $y$ denotes some statements. This table is may be used as definition of logical operator $\vee$. Note that $x\lor y \equiv y\lor x$ (so operator $\vee$ is associative).

So one may see that $x\lor\neg x = 1$. Indeed, according to $\vee$ truth table: $$ \begin{array}{cc|c} x & \neg x & \lor \neg x \\ \hline 1 & 0 & 1 \\ 0 & 1 & 1 \\ \end{array} $$ so $x\lor\neg x = 1$ regardless of whether $x$ is true or false.

In the same way one may see that $1\lor x = 1$ for bouth $x$ true and false.

Anton Grudkin
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  • so if i had to construct another English form sentence, which is logically equivalent to that one? – Mike Shasaco Jun 19 '16 at 19:28
  • @MikeShasaco there are so many ways to construct sentenses, so your question is too broad to be answered. Generally, use the Boolean algebra for simplifying sentenses in a such way (see here: https://en.wikipedia.org/wiki/Boolean_algebra). – Anton Grudkin Jun 19 '16 at 19:37
  • ok i get the truth table part but i was just a little confused on the aspect of making a logical equivalent to the sentence of "Susan goes to school or Susan does not talk on the phone or Susan does not go to school." – Mike Shasaco Jun 19 '16 at 19:43
  • @AntonGrudkin would "Susan's name is Susan" be logically equivalent since it is also true? – snulty Jun 21 '16 at 22:16
  • @snulty Yes, this sentence is logical equivalent to the sentence from the question since they bouth are true. Formaly we represent sentence like 'Susan's name is Susan' as one which is always true regardless of values of statements $A$ and $B$, and as "behavior" of sentence $A\lor B\lor \neg A$ is the same we say that these sentences are logically equivalent (by definition of logical equality). – Anton Grudkin Jun 22 '16 at 00:24
  • @MikeShasaco maybe if Anton added this example of logically equivalent sentence to the end of his nice answer, you would accept the answer? – snulty Jun 22 '16 at 11:31
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$A\text{ or } B \text{ or not }A$.

One can show by using truth tables that "or" is associative, so that the above is the same as

$\Big(A \text{ or not } A\Big) \text{ or } B.$

The statement $A\text{ or not }A$ is true regardless of whether $A$ is true or false. Hence the statement

$$ \Big(A \text{ or not } A\Big) \text{ or } B $$ must be true because $E \text{ or }F$ is true if either $E$ is true or $F$ is true.

Hence the given statement must always be true. And "always" means it is true regardless of whether Susan talks on the phone or not, and regardless of whether she goes to school or not. "Always" means "in all four cases". "All four cases" means "all four lines of the truth table". $$ \begin{array}{c|c|c} & & \text{Susan goes to school} \\ & & \text{or Susan does not talk on the phone} \\ \text{Susan talks on the phone} & \text{Susan goes to school} & \text{or Susan does not go to school.} \\ \hline T & T & T \\ T & f & T \\ f & T & T \\ f & f & T \\[6pt] \hline & & \uparrow \\[6pt] & & \text{All four} \\ & & \text{are true.} \end{array} $$

Michael Hardy
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