# Homework Help: Is there an easy way to learn Logic?

1. Oct 24, 2005

is there an easy way to learn Logic? Aside from all the funky looking squiggly symbols. I'm just not getting it as fast As I'd like to. I got a mid-term next month on Monday and I feel like I'm the only one that doesn't understand it. example:

1. (B.C)vD
2. ~C
3.Dv(B.C)
4.(DvB).(DvC)
5. DvC
6. D

i have to find the poroofs, and state the line or lines from which the argument is valid. Can someone direct me somehow

2. Oct 24, 2005

### AKG

I'm not exactly sure what's going on here. I guess this is supposed to be an argument, and the first n lines are supposed to be premises and the remaining lines are supposed to follow from the premises, where you have to figure out what n is? Then you have to justify each line?

Well go line by line. Is it possible that there are no premises? In general, this is possible. For example, you can prove things like P -> P without any premises, and this shouldn't surprise you. Also, any sentence is either true or false, so you can derive P v ~P from no premises. Such sentences are called theorems. One key thing about theorems is that they're also tautologies and vice versa, so they are true for any truth-value assignment of their atomic components. Another theorem is (((A -> B) -> A) -> A). Looks kind of complicated, but it makes for a better example. Look at its possible truth values:

A = T, B = T
A = T, B = F
A = F, B = T
A = F, B = F

Go through the calculations and check that the above sentence is true for any of these truth-value assignments.

Now, is sentence 1. (B.C)vD a theorem? You should be able to see by looking at it that it is not, but if you're not sure, pick a truth-value assignment on which it's false. What do you need for a disjunction to be false? Both its disjuncts must be false, so D must be false, and B.C must be false. What do you need for B.C to be false? At least one of B or C must be false. So if you choose them all to be false, you can see that this is a truth-value assignment on which (B.C)vD is false, so it's not a tautology, hence not a theorem. So you have to have at least sentence 1 in your premises.

Can you derive sentence 2 from sentence 1? You should be able to easily see that this is not the case. If you can't, then convince yourself by choosing a truth-value assignment on which (B.C)vD is true but ~C is false. What this tells you is that there's a case where 1. is true but 2. is false, so you certainly can't derive 2. from 1. Hopefully you can see easily that 1. is not a theorem and that 2. can't be derived from 1., and hopefully you don't find that you need to come up with these truth-value assignments, but if you need to do that to convince yourself for now, then do so. But can you see that if you know that (B.C) v D, there's no reason to believe C is false. If someone tells you, "either I will work out both my Biceps and my Calfs today, or I will work out my Delts" there's obviously not enough information there to conclude that I won't work out my calfs today. So you can tell that 1. doesn't entail 2. And you can certainly tell that 1. is not just true without any assumptions. Is it true a priori that I will either workout both my biceps and calfs, or else my delts? No, so 1. can't be true without some premises or without being a premise itself.

Anyways, so you know that you need at least 1. and 2. for your premises. Now can you derive 3. if you know 1. and 2.? 3. is just 1. with the order switched around. Certainly you can get 3. from 1. If someone tells you, "either A or B is true," then it's a real no-brainer to conclude that either B or A is true. The rule that allows you to do this is called commutativity. In general, the term "commute" refers to things that can switch places. Mutliplication is normally commutative. So a x b = b x a. Subtraction is not commutative, 5 - 3 = 2, but 3 - 5 = -2. Commuters are people who go back and forth to work every day, so they keep switching locations, so maybe this can help you remember that the rule that allows you to switch the places of disjuncts is called commutitivity. This rule also works for conjunction, A and B is the same as B and A, and that too is a no brainer. But it does not work for implication. If A -> B, that doesn't mean B -> A, and that too should be pretty clear.

So we have a hunch that we only need 1 and 2 as our premises. Note that just because we can derive 3 from 1 and 2, doesn't mean that we can derive the rest of the sentences from 1. and 2. Maybe 4 doesn't follow from the above, so we would need at least 1, 2, 3 and 4 as premises, and it just happens to be a stupid coincidence that you can derive one of the premises (3) from another (1). It makes 3 a superfluous premise, but that doesn't make 4 superfluous. But 4. does in fact follow from what's above it, and thus from 1 and 2. It follows from 3 (by a rule called distributivity) which follows from 1. Distributivity says that Av(B.C) = (AvB).(AvC), and there's also A.(BvC) = (A.B)v(A.C). It's similar to the distributive law for multiplication a*(b + c) = (a*b)+(a*c). So * distributes over +. It's not true the other way around, ie. a+(b*c) is not (a+b)*(a+c) in general. But with v and ., it goes both ways.

5 follows from 4 by conjunction elimination. This says that if "A and B are true" then you can always conclude "A is true" and "B is true". Finally, 6 follows from 5 and 2. 5 says that either D or C is true, so at least one of them is true. But 2 says that C is false, so D must be true. This uses a rule called hypothetical syllogism. It says that if you know (AvB) and ~B, you can conclude A.

3. Oct 25, 2005

### Jeff Ford

This chapter on Logic is easy to understand and well written

http://faculty.swosu.edu/michael.dougherty/book/chapter01.pdf" [Broken]

Last edited by a moderator: May 2, 2017