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
I need help solving this proof: 1. p ^ (~q) -> s 2. ~ s 3. p Conclusion: s -> q?
Proof :[Given] [math](p \wedge \neg q) \supset s[/math] [Given] [math]\neg s[/math] [Given] [math]p[/math] [1,2 modus tollens] [math]\neg (p \wedge \neg q)[/math] [4 tautology] [math]\neg p \vee q[/math] [3,5 resolution] [math]q[/math][Assumption] [math]\neg s[/math][6,7 [math]\vee[/math] intro] [math]q \vee \neg s[/math][8 tautology] [math]s \supset q[/math][7-9 [math]\supset[/math] intro] [math]\neg s \supset (s \supset q)[/math][2, 10 Modus ponens] [math]s \supset q[/math]The proof is quite simple. We are given that [math](p \wedge \neg q) \supset s[/math] and [math]\neg s[/math]. The only way an implication can be true if the consequent is false, is if the premise of the implication is false. This means that (line 4) [math]\neg (p \wedge \neg q)[/math] must hold. But since [math]\neg (p \wedge \neg q) \equiv \neg p \vee q[/math], line 5 follows. We are also given that [math]p[/math] so by resolution it follows that [math]q \vee \top[/math] which is equivalent to [math]q[/math] (line 6).If we further assume [math]\neg s[/math] (line 7) we can use or- introduction to yield [math]q \vee \neg s[/math] which by tautology is the same as [math]s \supset q][/math]. So assumption of [math]\neg s[/math] leads us to the conclusion that [math]s \supset q[/math], which means that [math]\neg s \supset (s \supset q)[/math]. But we are also given that indeed [math]\neg s[/math] holds, so one more inference give us the conclusion [math]s \supset q[/math].
I need help proving this proof: 1. p -> r v (~s) 2. s -> q 3. (s ^ q) -> ~s Conclusion: p -> (~s)?
I assume that you want to prove the following.Suppose [math]p \Rightarrow r \vee \neg s[/math] and [math]s \Rightarrow q[/math] and [math](s \wedge q) \Rightarrow \neq s [/math]We want to prove that [math]p \Rightarrow \neg s[/math]. One can show this by contradiction; suppose [math]p[/math] holds but that also [math]s[/math]. But then we get from (2) that we must have that [math]q[/math] holds. And then from (3) we get that [math]\neg s[/math] holds, which is a contradiction. Note that we do not use (1) here, and that makes me wonder if this is intentional.
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 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.
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
Solving formal proofs of validity for Logic class? Please Help?
I've been trying to solve these for a while now, if anyone can figure them out it would be much appreciated. _____________ Instructions: 1. Construct a formal proof of validity for the following arguments. 2. Attend carefully to instructions attached to specific arguments. 3. If a specific argument’s conclusion presents a “=” as its main connective, employ the parallel method of proof, recalling that alteration of the conclusion is restricted to the Rule of Replacement, which now includes Absorption (Abs.). For example, given the argument in our text, Exercise E-18 on page 417: J v (~J • K) J > L //therefore (L • J) = J The conclusion should be designated as //therefore C1 (L • J) = J when beginning the proof. At least one further step must be taken (at C2) to comply with the instructions. _____________________________ 1. Without Modus Ponens M > W M //therefore W _____________________________ 2. Without Exportation A > (D > K) //therefore (A • D) > K _____________________________ 3. No Rule Exceptions T = (L > K) ~ N • K //therefore T _____________________________ 4. No Rule Exceptions M = W //therefore M = (M • W) _____________________________ 5. All political projects offer evidence to the effect that if the people prosper then the people do not prosper. The people do not prosper. Key: P = The people prosper.