Natural Deduction Hurley, Logic 7.2.

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Presentation transcript:

Natural Deduction Hurley, Logic 7.2

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]

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

Practice Finding Proof Steps K → L (M → N) ● S N → T K v M / L v T M → N 2, Simp M → T 3,5, HS (K → L) ● (M → T) 1,6, Add L v T 4,7, CD

Practice Finding Proof Steps ~M ● N P → M Q ● R (~P ● Q) → S /S v T ~M 1, Simp ~P 2,5, MT Q 3, Simp ~P ● Q 6,7, Conj S 4,8, MP S v T 10, Add

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 3, Simp ~T ● A 1,4, MP ~T 5, Simp ~T v A 6, Add H → T 2,7, MP ~H 6,8, MT ~H ● ~T 6,9, Conj

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 3, 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 ;)

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 3, Simp ~S v B 4, Add S v K 1,5, MP K 4,6, DS K v ~D 7, Add H → S 2,8, MP ~H 4,9, MT

Practice Finding Proof Steps [A v (K ● J)] → (~E ● ~F) M → [A ● (P v R)] M ● U / ~E ● A M 3, Simp A ● (P v R) 2,4, MP A 5, Simp A v (K ● J) 6, Add ~E ● ~F 1,7, MP ~E 8, Simp ~E ● A 6,9, Conj

Practice Finding Proof Steps ~H → (~T → R) H v (E → F) ~T v E ~H ● D / R v F ~H 4, Simp ~T → R 1,5, MP E → F 2,5, DS (~T → R) ● (E → F) 6,7, Conj R v F 3,8, CD

Practice Finding Proof Steps (M v N) → (F → G) D → ~C ~C → B M ● H D v F / B v G M 4, Simp M v N 6, Add F → G 1,7, MP D → B 2,3, HS (D → B) ● (F → G) 8,9, Conj B v G 5,10, CD