Skip to main content

Subsection 3.3.8 The p ’s and q ’s are Placeholders

The identities that we have just defined (and inference rules that we are about to define) are sound ways of deriving new claims whose truth follows from the truth of our premises.

Since the job of the rules is to allow us to combine and modify well-formed formulas, we have needed a way to state them in terms of placeholders – slots that can be filled with whatever formulas we happen to be working with.

We’ve used variables such as p and q to do that. But, at proof time, we can substitute, for all such variables, any wffs.

The only thing that we must be careful about is that we must substitute uniformly. If one instance of a variable, say p , is replaced by some wff, say ( ab ), then every instance of p must be replaced by ( ab ).

Video cover image
Figure 3.3.2. Variable Substitution

Distributivity tells us that: \((p \wedge (q \vee r)) \equiv ((p \wedge q) \vee (p \wedge r)) \)

Suppose that we have:

[1] \((A \wedge B) \wedge (C \vee D) \)

Then substituting \((A \wedge B) \) for p, C for q, and D for r, we can used Distributivity to prove:

[2] \(((A \wedge B) \wedge C) \vee (A \wedge B) \wedge D)\)

Big Idea

When we state logical rules, variables are placeholders for arbitrary wffs.

Exercises Exercises

Exercise Group.

1. Recall that one form of De Morgan’s Laws is:

¬( pq ) ≡ ¬ p ∨ ¬ q

Consider: [1] ¬(( PQ ) ∧ ( RS ))

1.

(Part 1) To apply De Morgan to [1], we should let p equal which of the following:

  1. \(\displaystyle P \)

  2. \(\displaystyle Q\)

  3. \(\displaystyle (P \wedge Q)\)

  4. \(\displaystyle (R \wedge S) \)

  5. \(\displaystyle \neg (P \wedge Q) \)

  6. \(\displaystyle \neg ((P \wedge Q) \wedge (R \wedge S)) \)

Answer.
Correct answer is C.
2.

(Part 2) To apply De Morgan to [1], we should let q equal which of the following:

  1. \(\displaystyle P \)

  2. \(\displaystyle Q\)

  3. \(\displaystyle (P \wedge Q)\)

  4. \(\displaystyle (R \wedge S) \)

  5. \(\displaystyle \neg (P \wedge Q) \)

  6. \(\displaystyle \neg ((P \wedge Q) \wedge (R \wedge S)) \)

Answer.
Correct answer is C.
3.

(Part 3) What is the result of applying De Morgan once to [1]?

  1. \(\displaystyle \neg(P \wedge Q) \vee \neg (R \wedge S) \)

  2. \(\displaystyle \neg (P \vee Q) \vee \neg(R \wedge S) \)

  3. \(\displaystyle (\neg P \vee \neg Q) \vee (\neg R \vee \neg S)\)

  4. \(\displaystyle \neg (P \vee Q) \wedge \neg (R \vee S) \)

  5. \(\displaystyle (\neg P \vee \neg Q) \wedge (R \vee S) \)

Answer.
Correct answer is A.
Solution.
Explanation: Let p = \((P \wedge Q)\) and q = \((R \wedge S)\) .