How do you solve this logic proof?
I have 5 problems for logic... I have only learned these proofs A MPP MTT DN CP &I &E VI VE DS 1. (H->I)&(J->K),(LvK)->L,~L /- ~(HvJ) 2. (V->W)&(X->Y),(~W->Z)&(Y->~A),(Z->~B)&(~... /- ~B&C 3. /- (P->Q)->(~(Q&R)->~(R&P)) 4. --((A->~A)&(~A->A)) /- ~((A->~A)&(~A->A)) 5. (W->M)&(I->E),W v I, (W->~E)&(I->~M) /- E <-> ~M
Can someone solve this logic proof ~ ((A->C) v (B->C)) ├ ~ (A->C) & ~ (B->C)?
Travis Brauer is correct in saying that DeMorgan’s justifies the syntactic derivability (“⊦”) of ~ (A->C) & ~ (B->C) from ~ ((A->C) v (B->C)), and he’s also correct in suggesting that a truth table decomposition of the two expressions will show that semantic entailment (“⊧”) also applies.The misleading part of his answer, however, is that it obscures the difficulties inherent in the application of DeMorgan’s to the two different expressions of material implication involved. By simply substituting “X” for “(A->C)” and “Y” for “(B->C)” the meaning of those material conditionals is obscured.DeMorgan’s indeed can be applied to derive the conjunction of the negations of those conditionals from the negation of their disjunction (or conversely, deriving the negation of their disjunction from the conjunction of their negations). Truth table analysis also indeed does demonstrate semantic equivalence of the two expressions in your example.But there’s more at stake than just that demonstration of derivability and entailment within the system of propositional logic and its binary, Boolean semantics — precisely because material conditional expressions are involved in your example.To see why this is problematic, please review Terry Rankin's answer to How can we prove the implication rule of logic?
Help with my Logic Conditional proof? How do I solve these problems?
Help me with whichever question you know. Thank you so much! 1) -(P --> -Q) Ⱶ P&Q 2) -P & -Q Ⱶ -(P v Q) 3) -(P v Q) Ⱶ -P & -Q Thank you so much!
Help me solve this 7 step proof please!?
1.Ray QS is an angle bisector of angle PQR 2.Angle PQS is congruent to angle SQR 3.The measure of angle PQS equals the measure of PQR 4.The measure of angle PQS+ the measure of angle SQR equals the measure of angle PQR 5.The measure of angle PQS+ the measure of angle PQS equals the measure of PQR 6. 2 times the measure of angle PQS equals the measure of angle PQR 7.The measure of angle PQS equals 1/2 of the measure of angle PQR I just need the order of the steps... here they are A. Definition of angle bisector B. Distributive Property C. Angle Addition Postulate D. Given E. Division Property of Equality F. Definition of congruent angles G. Substitution Property of Equality Thank You!
Help me solve this logic proof using natural deduction?
Hi (I > E) > C, C > ~C /- I 1. (I > E) > C Premise 2. C > ~C Premise 3. ~C v ~C 2 Material Implication 4. ~C 3 Tautology 5. ~(I > E) 1,4 Modus Tollens 6. ~(~I v E) 5 Material Implication 7. ~~I & ~E 6 De Morgan's Law 8. ~~I 7 Simplification 9. I 8 Double Negation
Help solving formal proofs of validity for Logic class?
1. ~Q 2. (Q>S) > [(Q.S) > P] 3. Q v [(Q.S) v Q] 4. (Q>S) . P 5. Therefore P v (Q.S) 6. (Q.S) v Q D.S. on 3,1 7. Q.S D.S. on 6,1 8. Q Simp. on 7 9. Q . ~Q Conj. on 8,1 We cannot have both Q and ~Q. The argument is invalid. ____________________ 1. A>(B>H) 2. D v A 3. ~D 4. [(~E.H) > G] . [H > ~(B.H)] 5. (~E.H) v H 6. ~G 7. Therefore ~B 8. G v ~(B.H) C.D. on 4,5 9. ~(B.H) D.S. on 8,6 10. A D.S. on 2,3 11. B>H M.P. on 1,10 12. B>(B.H) Abs. on 11 13. ~B M.T. on 12,9 Q.E.D. As a matter of fact, you don’t even need premise 6 for an indirect proof: 8. A D.S. on 2,3 9. B>H M.P. on 1,10 10. B>(B.H) Abs. on 9 11. H>~(B.H) Simp. on 4 12. B>~(B.H) H.S. on 9,11 Prove by contradiction: 13. B (Assertion) 14. B.H M.P. on 10,13 15. ~(B.H) M.P. on 12,13 Contradiction 16. Therefore ~B Q.E.D.