Derivations
in [Logic]
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From: Fred on 20 Jun 2005 00:47 My professor has reduced the number of SD rules to eight, deleting the rules of biconditional and the rule of reiteration because they are redundant. BTW, there is no need to use vertical lines in proofs. And the rules should be written in metalogical symbols. Using double negation as the rule of negation elimination is better. Anyway, I will use your rule of negation elimination for demonstration. I will demonstrate some of your problems. C and F are left to you to exercise your skills. a) {A v B, ~B}├ A 1) A v B PREMISS 2) ~B PREMISS 3) A (ASSUMPTION) Discharged 4) B (ASSMP) Discharged 5) ~A (ASSMP) Discharged 6) B & ~B 2,4 CONJ. INTRO. 7) A 5-6 NEGATION ELIMINATION 8) A (3,4)-7 DISJUCTION ELIMINATION b) {A→(~B→C), A & ~B}├ C v E 1) A→(~B→C) Premiss 2) A & ~B Premiss 3) A 2, Conj. Elimination 4) ~B 2, Conj. Elimination 5) ~B→C 1, 3, Conditional Elimination 6) C 5, 4, Conditional Elimination 7) C V E 6, Disjuction Intro. c) {(~A v ~B)→C, D & ~C}├ A d) {A → ~~B, C → ~B}├ ~(A & C) 1) A → ~~B Premiss 2) C → ~B Premiss 3) A & C (Assumption) Discharged. 4) A 3, Conj. Elimination 5) C 3, Conj. Elimination 6) ~B 2, 5 Condictional Elimination 7) ~~B 1, 4 Condictional Elimination 8) ~B & ~~B 6, 7 Conj. 9) ~(A & C) 3-8 Negation Intro. e) A →(B → A) 1) A (Assumption) Discharged 2) B (Assumption) Discharged 3) B→A 2,1 Conditional Intro. (There is no need to introduce the rule of reiteration) 4) A→(B→A) 1,3 Conditional Intro. f) ~A → ((B & A) → C) g) (A v B) → (B v A) 1) A V B (Assumption) Discharged 2) A (Assumption) Discharged 3) B V A 2, Disjuction. Intro 4) B (Assumption) Discharged 5) B V A 4, Disj. Intro 6) B V A 2-5 Disjuction Elimination 7) (A v B) → (B v A) 1-6 Conditional Intro. h) A â†â†’ ~~A A) 1) A (Assumption) Discharged 2) ~A (Assumption) Discharged 3) A & ~A 1, 2 Conj. Intro 4) ~~A 2-3 Negation Intro. 5) A → ~~A 1-4 Conditional Intro. B) 1) ~~A (Assumption) Discharged 2) ~A (Assumption) Discharged 3) ~A & ~~A 1,2 Conj. Intro 4) A 2-3 Negation Intro. 5) ~~A → A 1-4 Conditional Intro. Combine Section A and B by the rule of bicondictional.
From: Fred on 21 Jun 2005 07:16 1st Semester Logic Student <jzarwel(a)gmail.com> wrote： >Hey! >Thanks for the help with those! If you have a sec, >could you help me >with a last five? These derivations are killing me! >They are: >a) |- (A > C) > ((B > C) > ((A v B) > C)) >b) |- (A > ~B) > (B > ~A) >c) |- ((A t B) & (B t C)) > (A t C) >d) {A > B, B > C, -C} |- ~A >e) {~A v B, A} |- B _______________________________________ 1st Semester Logic Student, These derivations aren't difficult to most students. I've demonstrated most of them. Try to discover the strategies within these proofs, then apply them in C. I have an extra problem for you. Try to prove {[(A → B)→ A]→ A} by using SD rules only. You can only learn logic by doing it. Try yourself. You'd get a good grade if you can solve it. a) ├(A → C) → [(B → C)→((A v B)→ C)] 1) A → C (Assumption) Discharged 2) B → C (Assumption) Discharged 3) A v B (Assumption) Discharged 4) A (Assumption) Discharged 5) C 1, 4 →E 6) B (Assumption) Discharged 7) C 2, 6 →E 8) C (4,6)-7 VE 9) (A v B)→ C 3-8 →I 10) (B → C)→((A v B)→ C) 2-9 →I 11) (A → C) → [(B → C)→((A v B)→ C)] 1-10 →I b) ├(A → ~B) → (B → ~A) 1) A → ~B (Assumption) Discharged 2) B (Assumption) Discharged 3) A (Assumption) Discharged 4) ~B 1, 3, →E 5) B & ~B 2, 4, &I 6) ~A 3-5, ~I 7) B → ~A 2-6, →I 8) (A → ~B) → (B → ~A) 1-7, →I c) ├((A â†â†’ B) & (B â†â†’ C))→(A â†â†’ C) d) {A → B, B → C, ~C}├ ~A 1) A → B Premiss 2) B → C Premiss 3) ~C Premiss 4) A (Assumption) Discharged 5) B (Assumption) Discharged 6) C 2, 5 →E 7) C & ~C 6, 3 &I 8) ~B 5-7 ~I 9) B & ~B 5, 8 &I 10) ~A 4-9 ~I e) {~A v B, A}├ B 1) ~A v B Premiss 2) A Premiss 3) ~A (Assumption) Discharged 4) ~B (Assumption) Discharged 5) A & ~A 2, 4 &I 6) B 4-5 ~E 7) B (Assumption) Discharged 8) B (3,7)-7 VE Fred
From: H. J. Sander Bruggink on 21 Jun 2005 08:53 Fred wrote: > My professor has reduced the number of SD rules to eight, deleting the > rules of biconditional and the rule of reiteration because they are > redundant. > > BTW, there is no need to use vertical lines in proofs. And the rules > should be written in metalogical symbols. > > Using double negation as the rule of negation elimination is better. I disagree with almost everything you write above. Well, I don't actually disagree, but I think it's matter of taste. For example, I personally don't like this double negation elimination rule at all... groente -- Sander
From: Fred on 21 Jun 2005 11:33
Sander, As you've said before, it's a matter of taste. Using double negation or Reductio ad absurdumas as the rule of negation elimination is the same in that both deductive systems are equalvant. Fred. H. J. Sander Bruggink wrote: > > > I disagree with almost everything you write above. Well, I don't > actually disagree, but I think it's matter of taste. For example, > I personally don't like this double negation elimination rule at > all... > > groente > -- Sander |