Predicate Logic Practice 2: The Four Quantifier Rules (Answers)

Difficulty: What the hell

Here are the answers to our second baby predicate-logic practice. This time the main point was to get used to the four quantifier rules—U.I., U.G., E.I., and E.G.—along with the irritating but necessary business of flagging.

If you made it through these without wanting to throw your book across the room, congratulations.

Exercise 1

1. (∀x)(Px ⊃ Qx)

2. (∃x)Px

∴ (∃x)Qx

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3. Pa         E.I. 2 (a is a new flagged name)

4. Pa ⊃ Qa   U.I. 1

5. Qa         modus ponens 3, 4

6. (∃x)Qx   E.G. 5

Exercise 2

1. (∀x)(Px ⊃ Qx)

2. (∀x)(Qx ⊃ Rx)

3. (∃x)Px

∴ (∃x)Rx

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4. Pa         E.I. 3 (a is a new flagged name)

5. Pa ⊃ Qa   U.I. 1

6. Qa         modus ponens 4, 5

7. Qa ⊃ Ra   U.I. 2

8. Ra         modus ponens 6, 7

9. (∃x)Rx   E.G. 8

Exercise 3

1. ¬(∃x)Px

2. (∃x)Qx

∴ (∃x)(Qx • ¬Px)

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3. (∀x)¬Px     Q.N. 1

4. Qa           E.I. 2 (a is a new flagged name)

5. ¬Pa         U.I. 3

6. Qa • ¬Pa   conjunction 4, 5

7. (∃x)(Qx • ¬Px)   E.G. 6

Exercise 4

1. (∀x)(Px ⊃ Qx)

2. (∀x)(Qx ⊃ Rx)

∴ (∀x)(Px ⊃ Rx)

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3. Let a be arbitrary.

4. Pa ⊃ Qa   U.I. 1

5. Qa ⊃ Ra   U.I. 2

6. Assume Pa

7. Qa         modus ponens 4, 6

8. Ra         modus ponens 5, 7

9. Pa ⊃ Ra   conditional proof 6–8

10. (∀x)(Px ⊃ Rx)   U.G. 9

That last one is the important one. It shows why flagging matters. You are only allowed to generalize from a because a was treated as arbitrary rather than as some special individual smuggled in from elsewhere.

Next time, we can either do a few more quantifier-rule exercises or move on to slightly nastier proofs that combine the quantifier rules with things like reductio and conditional proof. Predicate logic only gets more annoying from here, but at least now the annoyingness is beginning to take shape.