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2

Regarding step 11 : $\lnot q \rightarrow (q \rightarrow p)$ it is a tautology; so, it is provable. You are trying to prove it from assumptions; you can simplify it as follows : 1) $p$ --- assumed 2) $\lnot q$ --- assumed 3) $\lnot q \lor p$ --- Add.2 4) $q \rightarrow p$ --- Impl.3 5) $\lnot q \rightarrow (q \rightarrow p)$ --- C.P.2-4 Then ...


1

There's a quicker way than your approach. By the law of excluded middle, $p \lor \neg p$ If $p$, then ... If $\neg p$, then ... Remember, anything implies a true statement, and a false statement implies anything. edit: I'm just suggesting that you start with $p \lor \neg p$, which is either an axiom or derivable from $p \dashv \vdash \neg \neg p$, ...


4

The answer is pretty simple to derive using truth tables:


2

There are lots and lots of examples: start with obvious: $p \lor q \lor r \lor \mathtt{true}$, or alsmost obvious: $p \lor q \lor r \lor \neg p$, something with conjunction: $T_1(p) \land T_2(q) \land T_3(r)$ for any tautologies $T_1,T_2,T_3$ on one variable like $x \lor \neg x$, something yet more complicated $T_1(p,q) \land T_2(q,r) \land T_3(r,p)$ for ...


1

$(p \vee \lnot p) \wedge (q \vee \lnot q) \wedge (r \vee \lnot r)$.


3

It is in fact equivalent to $\lnot P\rightarrow Q$, which is also equivalent to $\lnot Q\rightarrow P$. So yes, it is indeed sufficient. However, in a formal logic system you should probably prove this using whatever axioms/rules of deduction you have.


0

Here's a 23 step OTTER proof from the 21 letter single axiom of Meredith CCCCCpqCNrNsrtCCtpCsp. -----> EMPTY CLAUSE at 11.77 sec ----> 7420 [hyper,2,7233] $F. Length of proof is 23. Level of proof is 17. ---------------- PROOF ---------------- 1 [] -P(C(x,y))| -P(x)|P(y). 2 [] -P(C(N(C(p,N(N(p)))),q)). 3 [] ...


0

I use Polish/Lukasiewicz notation. The rule of Negation elimination that I use says that from N$\beta$ having the same scope as an instance of K$\alpha$N$\alpha$ we can infer $\beta$. assumption 1 | KaNa assumption 2 || Nb 2, 1 Negation Elimination 3 | b 1-3 Conditional Introduction 4 CKaNab. As a more axiomatic ...


1

Let $S$ be the proposition that sales will go up, $E$ the proposition that expenses will go up, $H$ the proposition that the boss will be happy. Suppose we are given that $P$ is true, where $P = (S \wedge H) \vee (E \wedge \neg H)$. With the stated definitions, it then is sound to say it is impossible that the boss will be both happy and unhappy; $\neg(H ...


0

You can verify this by a cursory search: the deduction is $$ \frac{(S\land H)\lor (E\land \neg H)}{\neg(S\land E)}, $$ where $S::=$ "Sales will go up." $H::=$ "The boss will be happy." $E::=$ "Expenses will go up." This deduction is not sound. For example, $S\land E\land H$ is consistent with the antecedent but not with the conclusion.


-1

Surely you can have two different emotions about two different things at the same time? e.g. you could love dogs and hate cats simultaneously? This is the same kind of thing, because the event causes the emotion ABOUT THAT EVENT, but an overall emotion does NOT cause the event. You would be correct if the statement was 'either the boss will be happy so sales ...


1

Yes. We always have $$(A\land C)\rightarrow A.$$ And in your case we have as well $$A\rightarrow B.$$ So we get that $$(A\land C)\rightarrow B.$$


3

Yes: $$(A\rightarrow B \land C\rightarrow B)\implies [(A \land C)\rightarrow B] $$ So the forward direction holds, but not the $\Leftarrow$ direction. (I.e. we don't have $(A\rightarrow B \land C\rightarrow B)\iff [(A \land C)\rightarrow B] ).$ What we do have, in terms of a biconditional, is the following: $$(A\rightarrow B \land C\rightarrow B) \iff ...


0

It is no "and", instead is "or". You can phrase like: at most either $A$ or $C$ is sufficient for $B$. We have two ases. The first says if we have $A$, then $B$. The other says if we have $C$, the. $B$.


1

If we analyse what you wrote, namely $(\neg r \land \neg p) \to q$, then it says "$q$ is necessary for $r$ and $p$ to be false". Equivalently, "$r$ and $p$ being false is sufficient for $q$". That is: It is necessary that hiking be safe if the berries are not ripe and there are no grizzlies. It is sufficient that the berries are not ripe and there ...


0

The statement that the condition is necessary is an implication in one direction, and the statement that this condition is not sufficient is the negation of the implication in the other direction. Thus, we have the structure (safe $\to$ conditions) $\land \neg$ (conditions $\to$ safe). Fleshing this out gives the answer: $$ [p\to(\neg r \land \neg p)]\land ...


0

You aren't entirely correct in your translations. In your example of 27, you keep "will" in the same part of the sentence, but they should be part of the proposition that you're moving around. I know, this seems small and pedantic, but to be completely concise and accurate, it's an important detail. I will mention, however, that your translation is probably ...


7

DeMorgan's laws: $\sim (x \vee y) = (\sim x ) \wedge (\sim y)$ $\sim (x \wedge y) = (\sim x ) \vee (\sim y)$ In some notations where the overline indicates the complement, the rule is "break the line, change the sign": $\overline{x+y} = \overline{x} \ \overline{y}$ $\overline{xy} = \overline{x} + \overline{y}$ By only manipulating the right hand ...


5

I'll show you why the left-hand side and left hand side of the claim are contradictory: $$\begin{align} \left[p \wedge (q \lor r)\right] &\equiv \left[( \lnot q \lor \lnot p) \wedge (\lnot r \ \ \lor \lnot p)\right]\tag{1}\\ \\ \iff (p\land q )\lor (p \land r) &\equiv \lnot(q \land p) \land \lnot(r \land p)\tag{2}\\ \\ \iff (p\land q )\lor (p \land ...


3

The left hand statement is equivalent to $$(p\land q)\lor (p\land r)$$ Its negation is precisely the right hand statement. So this is indeed a self-contradictory proposition.


2

The negation of the goal has the following conjunctive normal form: $$ \lnot(((p \rightarrow q) \land ( q \rightarrow r) \land p) \rightarrow r) \equiv (\lnot p \lor q) \land (\lnot q \lor r) \land p \land \lnot r $$ I.e., you have four clauses: $$ \begin{array}{cl} A:& \{\lnot p, q\}\\ B:& \{\lnot q, r\}\\ C:& \{p\}\\ D:& \{\lnot r\} ...


2

Well, you have $p$ in the antecedent, and you have $p\rightarrow q$, and together, by modus ponens, you get $q$. Now, $q$, with the implication $q\rightarrow r$ give you $r$, again, using modus ponens. So the conjunction in the antecedent (i.e. the three conjuncts in the antecedent) imply $r$. Can you now use this to complete your assignment? Remember, ...


0

Your statement is almost correct. In mathematics "only if" would mean "the only path to promotion includes washing the boss's car". Don't forget that "washing the boss' car" may not be enough to get "a path to promotion".


0

p -> q [implication] wash car -> promotion...if (wash car) then (promotion) promotion (q) can also be gotten without washing the boss's car (maybe you deserved it) p <-> q [equivalence] wash car <-> promotion ...if and only if (wash car) then (promotion) which is the same as "You can get promoted only if you wash the boss's car".


1

$(p \rightarrow r) \lor (q \land \lnot r)$ is not equivalent to $((p \rightarrow r) \lor \lnot r) \land q$. $\lnot ( q \rightarrow r) \lor (p \rightarrow r)=\text{(...as you did...)}=(p \rightarrow r) \lor (q \land \lnot r)$ $=(\lnot p \lor r) \lor (q \land \lnot r)$ $=(\lnot p \lor (r \lor (q \land \lnot r))$ $=(\lnot p \lor ((r \lor q) \land (r \lor ...


1

In your approach, you made a mistake in the derivation on the third line, it should be: $$[\lnot (\lnot p \lor q) \lor \lnot (q \rightarrow r)] \lor (p \rightarrow r)$$ so that your fourth line becomes: $$[(p\land \lnot q) \lor (q \land \lnot r)] \lor (p \rightarrow r)$$ At this point, there isn't really much to be done except for a tedious case-by-case ...


1

HINT We have to rewrite 1) as : "Only if John chops down the tree, will he be a lumberjack" as : if John will be a lumberjack, then he chops down the tree. This one has the "logical form" : $p \rightarrow q$; thus, its converse ($q \rightarrow p$) will be : if John chops down the tree, then he will be a lumberjack, while its contrapositive ...


0

When we have an or statement either the left hand side or the right needs to be true for it to be true. In this case we have p on the left and $ (p \land q)$ on the right. But $ (p \land q)$ can only be true if p and q are both true. But if p is true we can immediately deduce that the or statement is true without knowing the value of q (since the LHS will be ...


3

From a Venn Diagram standpoint, all of $A$ plus any subset of $A$ will still just be $A$


2

Your step $p\wedge s$ is unjustified. I would go from $u\wedge s$ to $s$, then combine that $s$ with the earlier $p$ to get $p\wedge s$.


3

The conclusion is valid. And you were very careful along the way, which is why I was surprised that you jumped a couple steps when citing $p\land s$. From $u \land s$, you get $s$ by conditional elimination (simplification). Then you need to use conjunction-introduction to infer $p \land s$. Apart from that, just as important as writing the steps is ...


3

Your work is fine. You have not done anything wrong in your work. You just stopped short of determining if the proposition you end with (and hence the original proposition) is a tautology, contradiction, or contingency. Now we need to interpret the result, with the last line logically equivalent to the original proposition, we can look at the last line, $x ...


1

This is a great problem. I've long forgotten how fun this stuff can be. I believe the longer solution is correct because even though there are no bears and no berries, that is not sufficient for safety to be realized. I would read the correct solution as: "There is safety when there are no bears and no berries AND when other danger-causing variables ...


0

D - Some entrepreneurs are doctors - is the correct answer: Let's call the set of Women W, the set of Doctors D and the set of Entrepreneurs E. you need to demonstrate: E ∩ D ⊊ ∅. By 1. you have: W ⊆ E, while by 2. you have: ∅ ⊋ W ∩ D. So ∅ ⊋ W ∩ D ⊇ W ∩ D ⊇ E ∩ D. And by the last ...


4

(D) is the correct answer as:- (A). All women might not be doctors as it is not explicitly mentioned. (B). There might be doctors which are not women (men, transgender?). Since nothing is mentioned explicitly about the general populace of doctors, we cannot assume that there are only women doctors. (C). There might be entrepreneurs which are not ...


1

Let's make two cases. Case 1: Men/non-women must be considered. Then you have correctly found reasons why A,B and C cannot be true. Hence D is your only option, and is correct in my opinion as well. Case 2: Only women are considered. In this case, D is no less correct. However, C and B also becomes true since we are only considering the set of women, and ...


4

This is a syllogism of type Bocardo. Check the Venn diagram on the Wikipedia page


1

I believe D is the correct answer because all of the female doctors are entrepreneurs and some females are not doctors. In fact, in given statement #2, you can replace "women" with "entrepreneurs" and arrive exactly at answer D.


1

For hiking on the trail to be safe, it is necessary but not sufficient that berries not be ripe along the trail and for grizzly bears not to have been seen in the area. $$(\underbrace{(\lnot r \land \lnot p)\leftarrow q}_{\text{it is neccessary}}) \underbrace{\land}_{\text{and/but}} (\underbrace{\lnot((\lnot r\land \lnot p)\rightarrow q)}_{\text{it is ...


-2

For hiking on the trail to be safe, it is necessary but not sufficient that berries not be ripe along the trail and for grizzly bears not to have been seen in the area. This is a bit vague. Do ripe berries on their own really pose a threat to hikers? Or is it the combination of bears and berries that is dangerous? Unless this is some kind of trick ...


1

Your interpretation about insufficiency is correct. The attached part is to discount sufficiency; i.e., the converse is not true. If the original statement had only been "it is necessary that berries...", then your original solution would apply.


-2

No. For example, (A$\land$B$\lor$C) is not meaningful, even though both $\land$ and $\lor$ associate. If we have that for any binary operations B1, B2, for all x, y, z, ((xB1y)B2z)=(xB1(yB2z)), then we can drop parentheses as you've suggested. Thus, if we just have a semigroup, then we can drop all parentheses. But, if we have more than one binary ...


2

Yes they are because the first statement is false only when D is true and both H and S are false. The same holds for the second statement. Thus they both have same truth values hence equivalent.


5

$\phi_1 \rightarrow (\phi_2 \rightarrow (\phi_1 \land \phi_2))$ is a tautology. Thus, apply modus ponens twice ...


2

Step 1 : Definition of a valid argument An argument is valid if and only if its conclusion is never false while its premises are true. Step 2 : Building the truth table We follow the standard method, enumerating all possible binary cases for the sentence letters $p$ and $q$, and apply the usual boolean operators to find the truth values of the premises ...


2

You seem confused about the very definition of a valid argument: An argument is valid iff whenever its premises are true, so is its conclusion. In your approach, you took $p$ and $q$ to be the premises. However, this is not quite true. In the given form: $A$$\underline{B}$$C$ $A$ and $B$ are the premises, and $C$ is the conclusion. In the present ...


1

Certainly $\land$ and $\lor$ can be used meaningfully in this way. This is because we can prove (e.g. by truth tables, a straightforward exercise) that they are associative, i.e.: $$A \land (B \land C) \iff (A \land B) \land C \qquad A \lor (B \lor C) \iff (A \lor B) \lor C$$ so that we can argue that it doesn't really matter whether we use parentheses or ...


1

True, since by the definition of tautology: From wiki A formula of propositional logic is a tautology if the formula itself is always true regardless of which valuation is used for the propositional variables So we can see from the truth table that $p,q,\gamma$ are all true therefore it's a tautology. False, take $\alpha=p\vee\neg ...


0

Hint Rif to : Kenneth Rosen, Discrete mathematics and its applications (7th ed), page 35 : Exercise 10 asked you to show that the above formulae are tautologies using truth tables. Then : Exercise 12 : Show that each conditional statement in Exercise 10 is a tautology without using truth tables. We have to use the logical equivalences listed in ...


0

Once I read a classic example which I would like to share with you. A politician said......... p(If I Will win) q(I will cut down taxes by half) People will feel cheated only in the case when the politician actually won but taxes remained the same. In this case, q is FALSE but p is true so p->q become false. We are okay with the rest of cases/situation, ...



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