## Subsection5.1.11Substituting One Variable for Another

Notice that, both when we introduce arbitrary elements and when we introduce names for “the one”, we end up rewriting logical expressions and substituting one name for another.

[1] ∃x (Boy(x) ∧ Stupid(x))

Then we can name “the one” b* and write:

[1a] Boy(b*) ∧ Stupid(b*)

What we’ve done is to substitute b* for x. Notice that, when we do this, we must do it consistently within the entire scope of the quantifier that binds x. Within that scope, every instance of x (and nothing else) must be replaced by b*.

We need a concise notation for describing such substitutions. Let P be any wff and let x and y be any variables. Then we’ll write:

P(x/y)

Read this as, “P, with x (whatever it is) substituted in every place that a free (unbound) instance of y (whatever it is) appeared.

So, continuing with our example, let P be the wff:

[3] Boy(x) ∧ Stupid(x)

Then P(b*/x) yields:

[1a] Boy(b*) ∧ Stupid(b*)

You may now be saying, “Wait. I thought that you said that substitution happens to free (unbound) variables. Yet, in [1], the instances of x are bound by the existential quantifier ∃.” True. But we’re letting P be [3], the (nonstatement) wff that is inside the scope of ∃. So the instances of x are free and we’ll substitute for them. This will be important since, as in [1a], we want to derive statements that are quantifier-free.

We are now ready to see the new rules. Two deal with ∀ and two deal with ∃. Two (one for ∀ and one for ∃) remove the quantifiers (in favor of arbitrary or existential elements). The process of introducing the arbitrary or existential element is called instantiation (i.e., defining an instance). Two (again one each for ∀ and ∃) reapply the quantifiers. This process is called generalization (i.e., making a general statement from an instance).

### ExercisesExercises

#### 1.

Let P be the wff: Hungry(x) ∧ Thirsty(x).

What is P(Gerry/x)?

1. Hungry(x, Gerry) ∧ Thirsty(x, Gerry)

2. Hungry(Gerry) ∧ Thirsty(Gerry)

3. Hungry(Gerry) ∧ Thirsty(x)

Solution.

Explanation: We substitute Gerry for every occurrence of x.

#### 2.

Let P be the wff: Friendly(x) ∧ Likes(x, y).

What is P(Sunny/x)?

1. Friendly(Sunny) ∧ Likes(y, Sunny)

2. Friendly(Sunny) ∧ Likes(x, y)

3. Friendly(Sunny) ∧ Likes(Sunny, y)

4. Friendly(Sunny) ∧ Likes(Sunny, Sunny)