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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:
but I could not understand this problem or the solution. Could anybody help me? |
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As it was suggested in a comment above, drawing a truth-table, especially when there are only two or three variables (i.e. atomic sentences) can really help to illustrate exactly when two given expressions are equivalent. In this case, we see that $(\sim p \lor q)$ is true exactly when $(p\rightarrow q)$ is true, and it is false exactly when $(p\rightarrow q)$ is false. That is, $(p\rightarrow q) \equiv (\sim p \lor q)$. Alternatively, we can recognize the equivalence of the two expressions simply by comparing column of truth-values corresponding to each expression and see that the two columns are identical, and hence, the expressions are logically equivalent. Logical Equivalence : (p → q) $\equiv$ (¬p ∨ q) $\;\;$ Note that $"\equiv"$ is equivalent to $"\iff"$ Another way to use the truth-table above is to see that the implication $(p \rightarrow q)$ is false if and only if the truth-value of $p$ is true and value of $q$ is false. Symbolically, we can express that fact by asserting that for the implication to be true, then (it cannot be the case) that $(p \land \sim q)$; in other words, $\sim (p \land \sim q)$. This conveys exactly the same information as the material implication $(p \rightarrow q)$. Note that $\sim (p \land \sim q) \equiv (\sim p \lor q)$, by De Morgan. As for understanding that when $(p \rightarrow q)$, then if $p$ is true, we must have that $q$ is true: perhaps the following analogy will help. In many respects, the proper inclusion (proper "is a subset of") relation corresponds to material implication, where $\subset$ corresponds to the $\rightarrow$ relation. For example, suppose $A \subset B$. Then if it is true that $x \in A$, then it must be true that $x \in B$, since $B$ contains $A$. However, if $x \notin A$, that does not mean that $x \notin B$, since if $A \subset B$, then $B$ contains elements that $A$ does not contain. |
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As Peter Suber (Philosophy Department, Earlham College) points out in his website titled “Paradoxes of Material Implication”, material implication is the price of truth-functionality. Here is the link: http://www.earlham.edu/~peters/courses/log/mat-imp.htm quote: “These are paradoxes in the ancient sense, violations of intuition. They are not contradictions. But, you may well ask, why would we adopt a type of implication with such counter-intuitive results? … Primarily, the answer is that we want a truth-functional kind of implication. Remember that a connective is truth-functional if we can figure out the truth-value of the statement solely on the basis of the truth-values of its components. If we use a truth-functional form of implications, then we can construct truth-tables for our implication statements” Regards, Mike Jones edit: I forgot to point out that material implication is not unknown in ordinary language. When I was about 6 years old, trying to catch one of the birds that lighted in our yard, my grandfather gave me some this joking advice: “If you want to catch a sparrow, all you have to do is put salt on its tail.” :-) further edit: Nonetheless, there is a tendency to avoid material implication. A good example is the Wikipedia article on absolute value, at the line: “If b > 0, two other useful properties concerning inequalities are:” (We need to digress a moment to point out that this line ought to be edited to say that b is non-negative. That is, the case where b = 0, although trivial, is still germane. In our discussion below, we will assume that such an edit has been done.) By material implication, it is irrelevant if b is non-negative, but, admittedly, relevance disappears if b < 0. Perhaps the chin-strokers at Wikipedia feel that they are making it easier to understand for the layman by filtering the “irrelevance”, but a heavy price is paid for that, namely, the disruption of the nice chart that they were creating. If you simply let material implication do its work, then you can continue with the chart in an uninterrupted way. I daresay that this example of spoon-feeding filtering of irrelevance is quite widespread, and may even be partly contributory to the difficulty in appreciating material implication when those moments come when it cannot be avoided. That is, if the unrequested filtering were not routinely done, then people would be used to mateial implication already, just as they are used to tricky idioms in ordinary language, complex word-play jokes, and so on. In other words, it is the needless UNFAMILIARITY with material implication that is partly to blame here. |
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$p \to q$ is only logically false if $p$ is true and $q$ is false. So if not-$p$ or $q$ (or both) are true, you do not have to worry about $p \to q$ being false. On the other hand, if both are false, then that's the same as saying $p$ is true and $q$ is false (De Morgan's Law), so $p \to q$ is false. Therefore, the two are logically equivalent. |
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Understanding that $p \implies q$ is equivalent to $\sim p \vee q$ is usually very tricky for most people first learning it. I learned it in a philosophy class, of all things. The basic idea is this: $p$ and $q$ are "statements" that are either True or False. Then $p \implies q$ is also a statement. That the first part of what's confusing, that we're talking about a "statement about statements." Example: "If you own a dog, then you own an animal." Here "you own a dog" is a statement that is either true or false, "you own an animal" is either true or false, and the whole sentence "if you own a dog, then you own an animal" is either true or false. Now most people would agree that the last statement, "if you own a dog, then you own an animal" is True since a dog is generally considered to be an animal. What does it mean for this statement to be True? Well, it means that if a person owns a dog, then he/she owns an animal. But it does not say anything about people who do not own a dog. For example, someone who owns a cat does own an animal, but our assertion has nothing to say about cat owners. In fact, this point is what causes a huge amount of trouble for novices in my experience. So all of this is my long-winded way of saying, that the statement "if you own a dog, then you own an animal" remains true even for people who are not dog owners. So let's now write $p \implies q$ as a truth table with what we've learned above. P --> T F Q -v ----------- T | T T F | F T In other words, our statement "if you own a dog, then you own an animal" is true provided that owning a dog means you own an animal. The statement is always true if you don't own a dog. And the only way for the statement to be false is to own a dog and yet somehow do not own an animal. In other words, the main thrust of an implication like $p \implies q$ is the assertion that if $p$ is true, we absolutely must have $q$ true. Now, what is another way to get this same truth table? $\sim p \vee q$ gives you the same table, so $p \implies q$ is equivalent to $\sim p \vee q$. |
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