In Exercises 13–32, determine whether the argument is valid or invalid. You may compare the argument to a standard form, given on page 155, or use a truth table.

x→y
∼ x
∴∼ y
there is suppose to be a line between x and the bottom y.

you can draw no conclusion here.

knowing that x --> y
tells you nothing about what ~x implies

To determine whether the argument is valid or invalid, we can use the method of a truth table.

We have the following statements:
1. x → y
2. ∼x

We need to determine the validity of the argument: ∴∼y

To do this, we will create a truth table with columns for x, y, x → y, and ∼y.

| x | y | x → y | ∼y |
|---|---|-------|----|
| T | T | T | F |
| T | F | F | T |
| F | T | T | F |
| F | F | T | T |

In this truth table, we can evaluate the statement ∼y based on the given conditions.

We find that in the third row, x is false (F) and ∼y is false (F). However, in this row, the argument is invalid because ∼y does not logically follow.

Therefore, we can conclude that the argument is invalid.

To determine whether the argument is valid or invalid, we can use the method of logical deduction. Let's break down the steps to get the answer.

The argument given is:

x → y
∼x
∴ ∼y

To evaluate the validity of this argument, we need to determine if the conclusion (∼y) logically follows from the premises (x → y and ∼x). If it does, then the argument is valid. If not, it is invalid.

One way to approach this is by using the method of a truth table to consider all possible combinations of truth values for x and y. Let's construct a truth table:

| x | y | x → y | ∼x | ∼y |
|---|---|-------|----|----|
| T | T | T | F | F |
| T | F | F | F | T |
| F | T | T | T | F |
| F | F | T | T | T |

In the truth table, the values for x → y are derived from the truth values of x and y. The value ∼x is the negation of x, and ∼y is the negation of y.

Now, let's evaluate each row in the truth table to determine if the premises (x → y and ∼x) are true (T) or false (F) and if the conclusion (∼y) matches the truth value.

In the first row, x is true (T), y is true (T), x → y is true (T), ∼x is false (F), but ∼y is false (F). The conclusion does not match, so this combination does not support the argument.

In the second row, x is true (T), y is false (F), x → y is false (F), ∼x is false (F), and ∼y is true (T). The conclusion matches, so this combination supports the argument.

In the third row, x is false (F), y is true (T), x → y is true (T), ∼x is true (T), but ∼y is false (F). The conclusion does not match, so this combination does not support the argument.

In the fourth row, x is false (F), y is false (F), x → y is true (T), ∼x is true (T), and ∼y is true (T). The conclusion matches, so this combination supports the argument.

Based on the truth table, we have two combinations that support the argument and two that do not. Since the argument is not valid in all cases, we conclude that the argument is invalid.

Thus, the argument "x → y, ∼x, ∴ ∼y" is invalid.