Presentation is loading. Please wait.

Presentation is loading. Please wait.

Natural Deduction Hurley, Logic 7.3.

Similar presentations


Presentation on theme: "Natural Deduction Hurley, Logic 7.3."— Presentation transcript:

1 Natural Deduction Hurley, Logic 7.3

2 Our Replacement Rules De Morgan’s rule (DM): ~(p ● q) :: (~p v ~q)
~(p v q) :: (~p ● ~q) Commutativity (Com): (p v q) :: (q v p) (p ● q) :: (q ● p) Associativity (Assoc): [p v (q v r)] :: [(p v q) v r] [(p ● (q ● r)] :: [p ● (q ● r)] Distribution (Dist): [p ● (q v r)] :: [(p ● q) v (p ● r)] [(p v (q ● r)] :: [(p v q) ● (p v r)] Double Negation (DN): p :: ~~p Remember that p, q, r, and s stand for any well-formed formula, no matter how complex. For instance, below is an example of DN: ~~(K v J) ● ~B __________ (K v J) ● ~B

3 Practice Finding Proof Steps
A → ~(B ● C) A ● C / ~B A , Simp ~(B ● C) ,3, MP ~B v ~C , DM C ● A , Com C , Simp ~~C , DN ~C v ~B , Com ~B ,8, DS

4 Practice Finding Proof Steps
D ● (E v F) (The long way…) ~D v ~F / D ● E D , Simp (E v F) ● D , Com E v F , Simp ~~D , DN ~F ,6, DS F v E , Com E ,8, DS D ● E ,9, Conj

5 Practice Finding Proof Steps
D ● (E v F) (The short way…) ~D v ~F / D ● E (D ● E) v (D ● F) , Dist (D ● F) v (D ● E) , Com ~(D ● F) , DM D ● E ,5, DS

6 Practice Finding Proof Steps
(G ● H) v (G ● J) (G v K) > L / L G ● (H v J) , Dist G , Simp G v K , Add L ,5, MP

7 Practice Finding Proof Steps
M v (N v O) ~O / M v N (M v N) v O , Assoc O v (M v N) , Com M v N ,4, DS

8 Practice Finding Proof Steps
K > (F v B) G ● K / (F ● G) v (B ● G) K ● G , Com K , Simp F v B ,4, MP G , Simp G ● (F v B) ,6, Conj (G ● F) v (G ● B) 7, Dist (F ● G) v (G ● B) 8, Com (F ● G) v (B ● G) 9, Com (F ● G) v (B ● G) (G ● F) v (B ● G) Com (G ● F) v (G ● B) Com G ● (F v B) Dist If you have a complex conclusion, and you’re stuck, try altering it using replacement rules to see if it suggests a solution

9 Practice Finding Proof Steps
~S / ~(F ● S) ~S v ~F , Add ~F v ~S , Com ~(F ● S) , DM

10 Practice Finding Proof Steps
J v (K ● L) ~K / J (J v K) ● (J v L) , Dist J v K , Simp K v J , Com J ,5, DS

11 Practice Finding Proof Steps
H > ~A A / ~(H v ~A) ~~A , DN ~H ,3, MT ~H ● ~~A ,4, Conj ~(H v ~A) , DM ~(H v ~A) (~H ● ~~A) DM

12 Practice Finding Proof Steps
R > ~B D v R B / D ~~B , DN ~R ,4, MT R v D , Com D ,5, DS

13 Practice Finding Proof Steps
(O v M) > S ~S / ~M ~(O v M) ,2, MT ~O ● ~M , DM ~M ● ~O , Com ~M , Simp

14 Practice Finding Proof Steps
Q v (L v C) ~C / L v Q (Q v L) v C , Assoc C v (Q v L) , Com Q v L ,4, DS L v Q , Com

15 Practice Finding Proof Steps
~(~E ● ~N) > T G > (N v E) / G > T ~~(E v N) > T , DM (E v N) > T , DN (N v E) > T , Com G > T ,5, HS

16 Practice Finding Proof Steps
H ● (C ● T) ~(~F ● T) / F ~~F v ~T , DM (H ● C) ● T , Assoc T ● (H ● C) , Com T , Simp ~T v ~~F , Com ~~T , DN ~~F ,8, DS F , DN

17 Practice Finding Proof Steps
~(J v K) B > K S > B / ~S ● ~J ~J ● ~K , DM ~J , Simp ~K ● ~J , Com ~K , Simp ~B ,7, MT ~S ,8, MT ~S ● ~J ,9, Conj

18 Practice Finding Proof Steps
(G ● H) v (M ● G) G > (T ● A) / A (G ● H) v (G ● M) , Com G ● (H v M) , Dist G , Simp T ● A ,5, MP A ● T , Com A , Simp

19 Practice Finding Proof Steps
~(U v R) (~R v N) > (P ● H) Q > ~H / ~Q ~U ● ~R , DM ~R ● ~U , Com ~R , Simp ~R v N , Add P ● H ,7, MP H ● P , Com H , Simp ~~H , DN ~Q ,11, MT

20 Practice Finding Proof Steps
~(F ● A) ~(L v ~A) D > (F v L) / ~D ~F v ~A , DM ~L ● ~~A , DM ~~A ● ~L , Com ~~A , Simp ~A v ~F , Com ~F ,8, DS ~L , Simp ~F ● ~L ,10, Conj ~(F v L) , DM ~D ,12, MT

21 Practice Finding Proof Steps
E > ~B U > ~C ~(~E ● ~U) / ~(B ● C) ~~E v ~~U , DM E v ~~U , DN E v U , DN (E > ~B) ● (U > ~C) ,2, Conj ~B v ~C ,7, CD ~(B ● C) , DM ~ (B ● C) ~B v ~C DM

22 Practice Finding Proof Steps
(J v F) v M (J v M) > ~P ~F / ~(F v P) (F v J) v M , Assoc F v (J v M) , Assoc J v M ,5, DS ~P ,6, MP ~F ● ~P ,7, Conj ~(F v P) , DM ~(F v P) ~F ● ~P DM

23 Practice Finding Proof Steps
~(K v F) ~F > (K v C) (G v C) > ~H / ~(K v H) ~K ● ~F , DM ~F ● ~K , Com ~F , Simp K v C ,6, MP ~K , Simp C ,8, DS C v G , Add G v C , Com ~H ,11, MP ~K ● ~H ,12, Conj ~(K v H) , DM ~ (K v H) ~K ● ~H DM

24 Practice Finding Proof Steps
(K ● P) v (K ● Q) P > ~K / Q v T K ● (P v Q) K , Simp ~~K , DN ~P ,5, MT (P v Q) ● K , Com P v Q , Simp Q ,8, DS Q v T , Add

25 Practice Finding Proof Steps
(T ● R) > P (~P ● R) ● G (~T v N) > H / H ~P ● (R ● G) , Assoc ~P , Simp ~(T ● R) ,5, MT ~T v ~R , DM (R ● G) ● ~P , Assoc R ● G , Simp R , Simp ~~R , DN ~R v ~T , Com ~T ,12, DS ~T v N , Add H ,14, MP

26 Practice Finding Proof Steps
B v (S ● N) B > ~S S > ~N / B v W (B v S) ● (B v N) , Dist (B > ~S) ● (S > ~N) ,3, Conj B v S , Simp ~S v ~N ,6, CD ~(S ● N) , DM (S ● N) v B 1, Comm B ,9 DS B v W , Add

27 Practice Finding Proof Steps
(~M v E) > (S > U) (~Q v E) > (U > H) ~(M v Q) / S > H Line 3 can give you ~M and ~Q … from each, with addition (adding E), you can get the antecedents of both 1 and 2. Modus Ponens will give you what you need to attain the conclusion by Hypothetical Syllogism

28 Practice Finding Proof Steps
~Q > (C ● B) ~T > (B ● H) ~(Q ● T) / B ~Q v ~T , DM [~Q > (C ● B)] ● [~T > (B ● H)] ,2, Conj (C ● B) v (B ● H) ,5, CD (B ● C) v (B ● H) , Com B ● (C v H) , Dist B , Simp

29 Practice Finding Proof Steps
~(A ● G) ~(A ● E) G v E / ~(A ● F) ~A v ~G , DM ~A v ~E , DM (~A v ~G) ● (~A v ~E) 4,5, Conj ~A ● (~G v ~E) , Dist ~A , Simp ~A v ~F , Add ~(A ● F) , DM ~(A ● F) ~A v ~F DM

30 Practice Finding Proof Steps
(M ● N) v (O ● P) (N v O) > ~P / N [(M ● N) v O] ● [(M ● N) v P] 1, Dist (M ● N) v O , Simp O v (M ● N) , Com (O v M) ● (O v N) , Dist (O v N) ● (O v M) , Com O v N , Simp N v O , Com ~P ,9, MP [(M ● N) v P] ● [(M ● N) v O] 3, Com (M ● N) v P , Simp P v (M ● N) , Com (P v M) ● (P v N) , Dist (P v N) ● (P v M) , Com P v N , Simp N ,16, DS

31 Practice Finding Proof Steps
(T ● K) v (C ● E) K > ~E E > ~C / T ● K [(T ● K) v C] ● [(T ● K) v E] , Dist (T ● K) v C , Simp C v (T ● K) , Com (C v T) ● (C v K) , Dist [(T ● K) v E] ● [(T ● K) v C] , Com (T ● K) v E , Simp E v (T ● K) , Com (E v T) ● (E v K) , Dist (K > ~E) ● (E > ~C) ,3, Conj (E v K) ● (E v T) , Com E v K , Simp ~E v ~C ,14 CD ~(E ● C) , DM ~(C ● E) , Com (C ● E) v (T ● K) , Com T ● K ,18, DS


Download ppt "Natural Deduction Hurley, Logic 7.3."

Similar presentations


Ads by Google