Natural Deduction Hurley, Logic 7.3

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

Practice Finding Proof Steps A → ~(B ● C) A ● C / ~B A 2, Simp ~(B ● C) 1,3, MP ~B v ~C 4, DM C ● A 2, Com C 6, Simp ~~C 7, DN ~C v ~B 5, Com ~B 5,8, DS

Practice Finding Proof Steps D ● (E v F) (The long way…) ~D v ~F / D ● E D 1, Simp (E v F) ● D 1, Com E v F 4, Simp ~~D 3, DN ~F 2,6, DS F v E 5, Com E 7,8, DS D ● E 3,9, Conj

Practice Finding Proof Steps D ● (E v F) (The short way…) ~D v ~F / D ● E (D ● E) v (D ● F) 1, Dist (D ● F) v (D ● E) 3, Com ~(D ● F) 2, DM D ● E 4,5, DS

Practice Finding Proof Steps (G ● H) v (G ● J) (G v K) > L / L G ● (H v J) 1, Dist G 3, Simp G v K 4, Add L 2,5, MP

Practice Finding Proof Steps M v (N v O) ~O / M v N (M v N) v O 1, Assoc O v (M v N) 3, Com M v N 2,4, DS

Practice Finding Proof Steps K > (F v B) G ● K / (F ● G) v (B ● G) K ● G 2, Com K 3, Simp F v B 1,4, MP G 2, Simp G ● (F v B) 5,6, 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

Practice Finding Proof Steps ~S / ~(F ● S) ~S v ~F 1, Add ~F v ~S 2, Com ~(F ● S) 3, DM

Practice Finding Proof Steps J v (K ● L) ~K / J (J v K) ● (J v L) 1, Dist J v K 3, Simp K v J 4, Com J 2,5, DS

Practice Finding Proof Steps H > ~A A / ~(H v ~A) ~~A 2, DN ~H 1,3, MT ~H ● ~~A 3,4, Conj ~(H v ~A) 5, DM ~(H v ~A) (~H ● ~~A) DM

Practice Finding Proof Steps R > ~B D v R B / D ~~B 3, DN ~R 1,4, MT R v D 2, Com D 2,5, DS

Practice Finding Proof Steps (O v M) > S ~S / ~M ~(O v M) 1,2, MT ~O ● ~M 3, DM ~M ● ~O 4, Com ~M 5, Simp

Practice Finding Proof Steps Q v (L v C) ~C / L v Q (Q v L) v C 1, Assoc C v (Q v L) 3, Com Q v L 2,4, DS L v Q 5, Com

Practice Finding Proof Steps ~(~E ● ~N) > T G > (N v E) / G > T ~~(E v N) > T 1, DM (E v N) > T 3, DN (N v E) > T 4, Com G > T 2,5, HS

Practice Finding Proof Steps H ● (C ● T) ~(~F ● T) / F ~~F v ~T 2, DM (H ● C) ● T 1, Assoc T ● (H ● C) 4, Com T 5, Simp ~T v ~~F 3, Com ~~T 6, DN ~~F 7,8, DS F 9, DN

Practice Finding Proof Steps ~(J v K) B > K S > B / ~S ● ~J ~J ● ~K 1, DM ~J 2, Simp ~K ● ~J 4, Com ~K 6, Simp ~B 2,7, MT ~S 3,8, MT ~S ● ~J 5,9, Conj

Practice Finding Proof Steps (G ● H) v (M ● G) G > (T ● A) / A (G ● H) v (G ● M) 1, Com G ● (H v M) 3, Dist G 4, Simp T ● A 2,5, MP A ● T 6, Com A 7, Simp

Practice Finding Proof Steps ~(U v R) (~R v N) > (P ● H) Q > ~H / ~Q ~U ● ~R 1, DM ~R ● ~U 4, Com ~R 5, Simp ~R v N 6, Add P ● H 2,7, MP H ● P 8, Com H 9, Simp ~~H 10, DN ~Q 3,11, MT

Practice Finding Proof Steps ~(F ● A) ~(L v ~A) D > (F v L) / ~D ~F v ~A 1, DM ~L ● ~~A 2, DM ~~A ● ~L 5, Com ~~A 6, Simp ~A v ~F 4, Com ~F 7,8, DS ~L 5, Simp ~F ● ~L 9,10, Conj ~(F v L) 11, DM ~D 3,12, MT

Practice Finding Proof Steps E > ~B U > ~C ~(~E ● ~U) / ~(B ● C) ~~E v ~~U 3, DM E v ~~U 4, DN E v U 5, DN (E > ~B) ● (U > ~C) 1,2, Conj ~B v ~C 6,7, CD ~(B ● C) 8, DM ~ (B ● C) ~B v ~C DM

Practice Finding Proof Steps (J v F) v M (J v M) > ~P ~F / ~(F v P) (F v J) v M 1, Assoc F v (J v M) 4, Assoc J v M 3,5, DS ~P 2,6, MP ~F ● ~P 3,7, Conj ~(F v P) 8, DM ~(F v P) ~F ● ~P DM

Practice Finding Proof Steps ~(K v F) ~F > (K v C) (G v C) > ~H / ~(K v H) ~K ● ~F 1, DM ~F ● ~K 4, Com ~F 5, Simp K v C 2,6, MP ~K 4, Simp C 7,8, DS C v G 9, Add G v C 10, Com ~H 3,11, MP ~K ● ~H 8,12, Conj ~(K v H) 13, DM ~ (K v H) ~K ● ~H DM

Practice Finding Proof Steps (K ● P) v (K ● Q) P > ~K / Q v T K ● (P v Q) K 3, Simp ~~K 4, DN ~P 2,5, MT (P v Q) ● K 3, Com P v Q 7, Simp Q 6,8, DS Q v T 9, Add

Practice Finding Proof Steps (T ● R) > P (~P ● R) ● G (~T v N) > H / H ~P ● (R ● G) 2, Assoc ~P 4, Simp ~(T ● R) 1,5, MT ~T v ~R 6, DM (R ● G) ● ~P 4, Assoc R ● G 8, Simp R 9, Simp ~~R 10, DN ~R v ~T 7, Com ~T 11,12, DS ~T v N 13, Add H 3,14, MP

Practice Finding Proof Steps B v (S ● N) B > ~S S > ~N / B v W (B v S) ● (B v N) 1, Dist (B > ~S) ● (S > ~N) 2,3, Conj B v S 4, Simp ~S v ~N 5,6, CD ~(S ● N) 7, DM (S ● N) v B 1, Comm B 8,9 DS B v W 10, Add

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

Practice Finding Proof Steps ~Q > (C ● B) ~T > (B ● H) ~(Q ● T) / B ~Q v ~T 3, DM [~Q > (C ● B)] ● [~T > (B ● H)] 1,2, Conj (C ● B) v (B ● H) 4,5, CD (B ● C) v (B ● H) 6, Com B ● (C v H) 7, Dist B 8, Simp

Practice Finding Proof Steps ~(A ● G) ~(A ● E) G v E / ~(A ● F) ~A v ~G 1, DM ~A v ~E 2, DM (~A v ~G) ● (~A v ~E) 4,5, Conj ~A ● (~G v ~E) 6, Dist ~A 7, Simp ~A v ~F 8, Add ~(A ● F) 9, DM ~(A ● F) ~A v ~F DM

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 3, Simp O v (M ● N) 4, Com (O v M) ● (O v N) 5, Dist (O v N) ● (O v M) 6, Com O v N 7, Simp N v O 8, Com ~P 2,9, MP [(M ● N) v P] ● [(M ● N) v O] 3, Com (M ● N) v P 11, Simp P v (M ● N) 12, Com (P v M) ● (P v N) 13, Dist (P v N) ● (P v M) 14, Com P v N 15, Simp N 10,16, DS

Practice Finding Proof Steps (T ● K) v (C ● E) K > ~E E > ~C / T ● K [(T ● K) v C] ● [(T ● K) v E] 1, Dist (T ● K) v C 4, Simp C v (T ● K) 5, Com (C v T) ● (C v K) 6, Dist [(T ● K) v E] ● [(T ● K) v C] 4, Com (T ● K) v E 8, Simp E v (T ● K) 9, Com (E v T) ● (E v K) 10, Dist (K > ~E) ● (E > ~C) 2,3, Conj (E v K) ● (E v T) 11, Com E v K 13, Simp ~E v ~C 12,14 CD ~(E ● C) 15, DM ~(C ● E) 16, Com (C ● E) v (T ● K) 1, Com T ● K 17,18, DS