# Appendix: Practice Problems

## First-Order Derivations:

This first set of derivations are all *syllogisms*: simple two-premise arguments whose natural language equivalents have been known to be valid for quite a long time.

exercise 0.1

0.1 Ax(F(x) -> G(x)), Ex(H(x)/\-G(x)) :|-: Ex(H(x)/\-F(x))

exercise 0.2

0.2 Ex(F(x)/\ -G(x)), Ax(F(x) -> H(x)) :|-: Ex(H(x)/\ -G(x))

exercise 0.3

0.3 Ax(F(x)->G(x)), -Ex(G(x)/\H(x)) :|-: -Ex(H(x)/\F(x))

exercise 0.4

0.4 Ax(H(x)->F(x)), -Ex(F(x)/\G(x)) :|-: -Ex(H(x)/\G(x))

exercise 0.5

0.5 -Ex(F(x)/\G(x)), Ex(H(x)/\F(x)) :|-: Ex(H(x)/\-G(x))

This set of derivations cover the quantifier equivalents of DeMorgan’s laws:

exercise 0.6

0.6 :|-: Ax(F(x)/\G(x)) <-> AxF(x)/\AxG(x)

exercise 0.7

0.7 :|-: Ex(F(x)\/G(x)) <-> ExF(x)\/ExG(x)

This set of derivations shows some implications similar to the above, which are valid, but which are not reversible (while the premise implies the conclusion, the conclusion does not imply the premise)

exercise 0.7

0.7 AxF(x)\/AxG(x) :|-: Ax(F(x)\/G(x))

exercise 0.8

0.8 Ex(F(x)/\G(x)) :|-: ExF(x)/\ExG(x)

We can also explore how the quantifiers can “travel over” or “commute” with other connectives (the second one here might be a little tricky. Try using the QN rule):

exercise 0.9

0.9 Ax(F(x)<->G(x)) :|-: AxF(x)<->AxG(x)

exercise 0.10

0.10 ExF(x)->ExG(x) :|-: Ex(F(x)->G(x))

## Truth Tables

Here are some truth table problems to practice on as well

exercise 0.11

0.11 :|-: ((P/\Q)\/R)<->((P\/R)/\(Q\/R))

exercise 0.12

0.12 :|-: P\/Q<->P/\Q

exercise 0.13

0.13 P\/R :|-: ~P->~~R

exercise 0.14

0.14 P :|-: P/\~(P->P)

exercise 0.15

0.15 ~P :|-: P<->~(P->P)

exercise 0.16

0.16 P->Q :|-: Q->P

exercise 0.17

0.17 P->Q :|-: ~Q->~P

exercise 0.18

0.18 P->Q, Q<->R :|-: P->R

exercise 0.19

0.19 ~P, P<->R :|-: R<->~(P->P)

exercise 0.20

0.20 P/\Q<->P\/R :|-:(P/\Q)\/~(P\/R)