I need to solve the following logical equivalence:   (p ↔  q) ↔ ((p∧q)∨(¬p∧¬q)) I know i need to solve the problem in 2 parts, and i think i can do it with raa but im stuck. So  if someone could give me some help it would be great. Thanks
124057
I am working on a problem where I need to show the logical equivalence of two propositions.  One is a biconditional: $p \leftrightarrow q$.  And the other is this: $(p \land q) \lor (\lnot p \land \lnot q)$  I can get the biconditional down to this: $(\lnot p \lor q) \land (\lnot q \lor p)$  And I don't know how to change the second proposition.  Can someone give me a place to start?
122834
This question comes from from my algebra paper:  $(p \rightarrow q)$ is logically equivalent to ... (then four options are given).   The module states that the correct option is $(\sim p \lor q)$. That is:      $$(p\rightarrow q) \iff (\sim p \lor q )$$   but I could not understand this problem or the solution. Could anybody help me?
11482
Using the Fitch system, how do I prove $((p \implies q) \implies p)\implies p$? I started with the hypotheses $(p \implies q) \implies p$ and $\sim p$. However, from these hypotheses I did not get the desired contradiction to solve the question.  Thanks sb45
46905
So I am studying computer science and right now I am stuck on a problem.  When does s∗ = s, where s is a compound proposition?  So far the only thing I can come up with is:  s* = s when the compound proposition is composed only of the same           propositions. (ex. p ∧ p = p ∨ p)  The book defines duality as:  The dual of a compound proposition that contains only the logical operators ∨, ∧, and ¬ is the compound proposition obtained by replacing each ∨ by ∧, each ∧ by ∨, each T by F, and each F by T. The dual of s is denoted by s∗. (Discrete Mathematics and its Applications, Rosen, 7e)  Any help would be great, this is a tricky one.
49089
If the union of two theories is not consistent there exists a sentence $P$ for which one of them implies $P$ and the other one implies $\lnot P$.  Can someone please check if this is true or not?  I used to prove this by contradiction. I assumed for every sentence $P$, if $T_1$ implies $P$, then $T_2$ satisfies $P$ or $T_2$ doesn't imply $P$ but never $T_2$ implies $\lnot P$. So i think there exists a model of $T_2$ as a substructure of a model of $T_1$( i'm not sure if this is true... I just say it intuitively ), so I take that model of $T_1$ as a model of  $T_1 \cup T_2$. Since the union doesn't have a model so there exists a sentence $P$ such that $T_1$ implies $P$ and $T_2$ implies the $\lnot P$.
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Let's say that I prove statement $A$ by showing that the negation of $A$ leads to a contradiction.   My question is this: How does one go from "so there's a contradiction if we don't have $A$" to concluding that "we have $A$"?  That, to me, seems the exact opposite of logical. It sounds like we say "so, I'll have a really big problem if this thing isn't true, so out of convenience, I am just going to act like it's true".
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The question gives the following: ¬(p ∨ ¬q) → (¬r ∧ (p → r))  With my own working, I have come to the answer of (p ∨ ¬q) ∧ (¬r ∧ ¬p) ∧ (¬r ∧ r). Can anyone tell me if this is correct?  Also, please tell me what Disjunctive Normal Form really means (without complicated terminology). The answer I got was from trying to break down what other similar questions have done. I still don't know what it really means.
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Suppose you have:      $x=(p,q)$ and $y=(p',q')$ with $p, q \in \mathbb{Z}$ and $q \not = 0$ then $x \sim y$ if $pq'=p'q$ is an equivalence relation?   So I managed to show that its reflexivity and symmetry, but I am unable to prove that it's transitive.
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$((\neg \alpha \to \alpha) \to \alpha) $  Hi, I am trying to prove this.  Can someone gives me some hints to start the question... My friend told me I might need to use deduction theorem here, but I have no clue to approach it..  Axiom 1: A→(B→A).  Axiom 2: (A→(B→C))→((A→B)→(A→C)).  Axiom 3: (¬B→¬A)→((¬B→A)→B).  To clarify: A, B, C, α, and β are propositions (i.e. assigned True or False). → and ¬ have the standard logical meanings.
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This semester I take Introduction to logic and sets theory and for logic our reference is A Mathematical Introduction to Logic by H.B.Enderton.  I have some problems with this course I mean it is somehow unfamiliar ! our professor teach well and I learn at class but when I want to solve my homework ,I have no idea ,how to solve problems !  I don't have any idea how to prove this :  suppose a and b are tautologically equivalent, where A is wff and a is used in A and B is  another wff that b is used in; prove that A and B are tautologically equivalent.  Any one can help me ?
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Why is "$P \Rightarrow Q$" equivalent to "$\neg Q \Rightarrow \neg P$"?  I am trying to understand why this does always apply, in terms of pure logic. Can you please explain it to me?
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Using only rules of natural deduction, I am trying to prove that $$p \implies q \vdash \lnot p \lor q$$ but am having a lot of difficulty. I was able to prove the other direction.   Could anyone show me or point me to the right direction?
110903
I'm having a bit of trouble proving this with the propositional calculus rules.  $$p \lor (p\implies q)$$ Would someone mind helping me and showing which rules they've used with an explanation! Thanks!
88806
Inference rules allowed are: $\lor$, $\wedge$, $\rightarrow$, $\neg$ introductions and eliminations.  In question, it says I am allowed to make use of a lemma or equivalence as long as I provide a proof of it by natural deduction too. (I will come to that soon enough, hopefully.)  First of all, that "$,$" between $p$ and $p \rightarrow (r \rightarrow q)$ made me a bit confused, but I ended up assuming that comma just splits two premises we are given. I tried to not dig too much and proceeded to make proof, however I got stuck after using implication elimination on $p \rightarrow (r \rightarrow q)$ and obtaining $(r \rightarrow q)$. From there on, I have a sense that I will just use 4-5 more steps and reach $(\neg q \to \neg r)$ but I just couldn't figure it out.  Then I realized that I can make use of equation $p \rightarrow q \equiv \neg p \lor q $ .  As stated in question, I tried to prove this equation by natural deduction (I know at this point I am just making things harder than it is supposed to be, but what the heck..)  This is the proof I tried to make in fitch style:    If you realize I am using double negation here and I am not even sure if that is even allowed to use.  From here, I just needed to show that $(r \rightarrow q) \equiv (\neg q \rightarrow r)$ knowing $(\neg r \lor q) \equiv (\neg \neg q \lor \neg r)$ and that is $(\neg r \lor q) \equiv (q \lor \neg r)$   But if you realize, that means I will have to prove double negation law "$\neg \neg q \equiv q$" as well as commutative law "$(\neg r \lor q) \equiv (q \lor \neg r)$". Like these extra two small proofs are not enough, I also have to use them with proper notation/syntax in latex aside with fitch style proof tables.  This makes me realize that it is definitely a bad idea to make this proof by natural deduction using these "transformation" equivalences if I may call them that.   So, I think I need to give up on that idea and stick to direct proof with implications.  However, as stated above, I am stuck after implication elimination. Could you guys help me figure it out how to move forward from there?  P.S: Reason I state all those unnecessary steps is that, I think it is just more appropriate to not just "ask for a direct answer" but provide my thought process. Another challenge of doing this proof for me is that having to use fitch style proofs in latex. I just can't figure out where to place that "p" before comma in the question. Anyway, enough confusion!
120002
I have several of these types of problems, and it would be great if I can get some help on one so I have a guide on how I can solve these.  I tried asking another problem, but it turned out to be a special case problem, so hopefully this one works out normally.   The question is:  Prove $(A - B) - C = A - (B \cup C)$  I know I must prove both sides are not equivalent to each other to complete this proof. Here's my shot: We start with left side.   if $x \in C$, then $x \notin A$ and $x \in B$. So $x \in (B \cup C)$ So $A - (B \cup C)$   Is this the right idea? Should I then reverse the proof to prove it the other way around, or is that unnecessary? Should it be more formal?  Thanks!
98084
How can I prove the Boolean algebraic rule $$A+A'B=A+B$$ without using a truth table?  With the truth table, it is easy to see that the two are equal, but how can I prove it using lesser Boolean identities?
107211
I'm given the following expresssion: $$ \forall a [\phi(a) \to \psi(a)] \wedge \forall a [\psi(a) \to \phi(a)] $$ that I wish to logically reduce to: $$ \forall a [\phi(a) \leftrightarrow \psi(a)] $$  The only area I'm uncertain about is showing that universal quantification is distributive over conjunction, as trivial as it seems. Rather, I'm not sure where to find theorems or lemmas related to the topic.  The set theory text from which this comes provides no insight and the introductory logic text I own also doesn't touch on the matter, at least in any great detail.