Decide whether statement (3) is true if statements (1) and (2) are both true.

discrete mathematicslogicpropositional-calculussolution-verification

I am trying to solve the following problem from the book Reading, Writing, and Proving: A Closer Look at Mathematics 2nd ed. on page 44 Problem 4.20)b)


Decide whether statement $(3)$ is true if statements $(1)$ and $(2)$ are both true. Give reasons for your answers.

The three statements are:

$(1)$ If Susie goes to the ball in the red dress, I will stay home.

$(2)$ Susie went to the ball in the green dress.

$(3)$ I did not stay home.


The following is my attempted solution:


The universe for all variables is the set of all humans.

Let $B(x)$ denotes the statement: $x$ goes to the ball.

Let $R(x)$ denotes the statement: $x$ wears a red dress.

Let $S(x)$ denotes the statement: $x$ stays home.

Each of the statement $(1),(2)$ is expressed symbolically as:

\begin{align}
&\exists x,(B(x) \wedge R(x)) \rightarrow \exists y,(y \ne x \wedge S(y)). &\text{(1)}\\
&B(x) \wedge \lnot R(x) &\text{(2)}
\end{align}

Because the statement $(2)$ makes the statement $(1)$ vacuously true, $(3)$ is True or False and not both.


I am not sure whether my answer is correct or not.

Reference:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 44). New York: Springer.

Best Answer

The third statement does not follow from the first two. If "Susie goes to the ball in the red dress" is A, and "I will stay home" is B, then of course $\bar{B}\implies \bar{A}$, where the bar denotes negation. But you cannot deny the hypothesis to conclude that $B$ is False, so $\bar{B}$ cannot be deduced from $\bar{A}$. The wording of this question makes it appear as if you should answer it with either "(3) is false" or "(3) is true", but I don't believe this is what the author intended.

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