1. Natural Deduction
Hurley, Logic 7.2

2. Our Next 4 Argument Forms
Constructive Dilemma (CD): (p → q) ● (r → s) p v r _______ q v s Simplification (Simp): p ● q p
Conjunction (Conj): p q
_______
p ● q
Addition (ADD)
p v q
Remember that p, q, r, and s stand for any well-formed formula, no matter how complex. For instance, below is an example of Addition:
~(K v J) ● ~B
__________
[~(K v J) ● ~B] v [(R v ~S) ● ~B]

3. Practice Finding Proof Steps
1. A → B 2. (B v C) → (D ● E) 3. A /D 4. B 1,3, MP 5. B v C 4, ADD 6. D ● E 2,5, MP 7. D 6, Simp

4. Practice Finding Proof Steps
K → L
(M → N) ● S
N → T
K v M / L v T
M → N , Simp
M → T ,5, HS
(K → L) ● (M → T) ,6, Add
L v T ,7, CD

5. Practice Finding Proof Steps
~M ● N
P → M
Q ● R
(~P ● Q) → S /S v T
~M , Simp
~P ,5, MT
Q , Simp
~P ● Q ,7, Conj
S ,8, MP
S v T , Add

6. Practice Finding Proof Steps
F → (~T ● A) This is problem 14 in 7.2, III
(~T v A) → (H → T)
F ● O / ~H ● ~T
F , Simp
~T ● A ,4, MP
~T , Simp
~T v A , Add
H → T ,7, MP
~H ,8, MT
~H ● ~T ,9, Conj

7. Practice Finding Proof Steps
(S v B) → (S v K) This is problem 15 in 7.2, III (Hurley 12th Edition)
(K v ~D) → (H → S)
~S ● W / ~H
~S , Simp
It appears the book made a mistake here (one clue is, it left off the first parenthesis on line 2)… after line 4, I believe this can’t proceed. I’ve included this as a lesson: if you get stuck, move on after a while ;)

8. Practice Finding Proof Steps
(~S v B) → (S v K) Again, problem 15 in 7.2, III (but Hurley 11th Edition)
(K v ~D) → (H → S)
~S ● W / ~H
~S , Simp
~S v B , Add
S v K ,5, MP
K ,6, DS
K v ~D , Add
H → S ,8, MP
~H ,9, MT

9. Practice Finding Proof Steps
[A v (K ● J)] → (~E ● ~F)
M → [A ● (P v R)]
M ● U / ~E ● A
M , Simp
A ● (P v R) ,4, MP
A , Simp
A v (K ● J) , Add
~E ● ~F ,7, MP
~E , Simp
~E ● A ,9, Conj

10. Practice Finding Proof Steps
~H → (~T → R)
H v (E → F)
~T v E
~H ● D / R v F
~H , Simp
~T → R ,5, MP
E → F ,5, DS
(~T → R) ● (E → F) ,7, Conj
R v F ,8, CD

11. Practice Finding Proof Steps
(M v N) → (F → G)
D → ~C
~C → B
M ● H
D v F / B v G
M , Simp
M v N , Add
F → G ,7, MP
D → B ,3, HS
(D → B) ● (F → G) 8,9, Conj
B v G ,10, CD