Symbolic Logic

Has anyone here taken a symbolic logic class?

I'm in one now and I must say, I've become a bit of a nerd for it. If anyone has taken one, describe the syntax you used, e.g. connectives such as conjunctions, disjunctions, conditionals, bi-conditionals, etc. Also, what rules of inference did you use for your formal proofs?

I've googled stuff on symbolic logic and found that there are systems that use completely different symbols for the connectives and have even more rules. As far as propositional logic goes, I've been utilizing the following 16 rules:

Simplification
Conjunction
Disjunctive syllogism
Addition
Modus ponens
Modus tollens
Conditional proof
Biconditional rule
Biconditional modus ponens
Biconditional modus tollens
Negated conjunction
Demorgan's law
Negated conditional
Negated biconditional
Double negation (composition/decomposition)
Indirect proof

Anyway, if anyone is familiar with this stuff, explain your system. Otherwise, I expect this thread to fade into oblivion...

Allow me to bore the shit out of you all...

Basically, in the beginning, you take simple English statements and learn how to translate them into what is called "L." Each proposition is assigned a capital English letter. For example, "Alice goes to the party" might be represented by the letter A where "Betty goes to the party" might be represented by the letter B. Then you look at the various ways of connecting these two statements. One could say, "Alice and Betty go to the party," which is another way of saying, "Alice goes to the party and Betty goes to the party."

In "L," this becomes: A & B
(A and B) Conjunction

"(Either) Alice or Betty goes to the party" becomes: A v B
(A or B) Disjunction

"If Alice goes to the party, then Betty will go (to the party)": A --> B
(If A, then B) Conditional

"Alice will go to the party if and only if Betty goes (to the party): A <--> B
(A if and only if B) Biconditional

Then you have the negation symbol: ~

Long story short, the 16 rules mentioned in the above post are ways by which statements in L, such as the ones above, can be constructed or deconstructed.

For example, simplification is a deconstructive rule that applies to conjunctions. When you have A & B, you can pull off either side. Conversely, if you have A and you have B on separate lines of a formal proof, you can apply the rule of conjunction to make A & B on the next line. What would be the purpose of doing this? Well, when you're looking at a formal proof, in order to prove validity, you have to use these rules in order to derive the conclusion from the given premises.

For example:

1. A & B
2. A --> C ........................................................Conclusion: C


Obviously, from premise 2, if we have A, then we get C. We have A. However, it is currently joined with B in premise 1. In order to utilize A, we need A by itself. The rule of simplification allows us to extract A (or B if we need it) from the conjunction. This would look like:

1. A & B
2. A --> C ........................................................Conclusion: C
3. A ................................................................1 simp

And we write over on the right side the number where we got the A and an abbreviation of how we derived it. Obviously, we got the A from premise 1, by the rule of simplification.

The next step concerns conditionals. The part of the conditional before the arrow is called the antecedent and the part after the arrow is the consequent. The rules for pulling apart conditionals are called Modus Ponens and Modus Tollens. In this case, we want C, so we are concerned with Modus Ponens. Modus Ponens is simple. It just says that if you have the antecedent, then you can derive the consequent:

1. A & B
2. A --> C ........................................................Conclusion: C
3. A ................................................................1 simp
4. C ................................................................2, 3 MP

Step 4 comes from numbers 2 and 3 by the rule of Modus Ponens (MP). And we're done since we arrived at the conclusion.

I'll stop here. This is just a quick intro for anyone interested (probably no one).

At the very least, it is a system to show how logic works mathematically. Symbolic logic has nothing to do with "symbolism" as you're referring to it. Its purpose is not to "point out" anything inherent in the world. Rather, it's about translating statements into logic, into math. And sometimes so-called common sense is incorrect.

As EverReady points out, symbolic logic can count as a math credit. For me, it can either count as philosophy or as math. In my opinion, it is much more fun and engaging than a standard math class, especially when it comes to formal proofs. There isn't necessarily one way to solve a formal proof. It's all about anticipating what moves are going to benefit you down the road. Some arguments are simple enough that you can catch if there is a fallacy just from listening to the English. Considering the example I gave a couple posts ago, if you apply English statements to the symbols, you might have something like:

1. Jason lives in Sacramento and he attends UC Davis.
2. If Jason lives in Sacramento, then he lives in California.

Now, obviously, you don't need to do any simplification move in your head to come to the conclusion that Jason lives in California. So my above example was an overly simple one. When the statements get more complicated, it is easier to deal with their logical structure in "L" form.

You may live your life just fine without it. Same thing with something like Calculus. I enjoy symbolic logic at the moment similar to how someone might enjoy Sudoku. However, I can understand how being able to translate arguments into their essential logical form can be helpful in analyzing whether the argument is valid or invalid. In this way, there is more application for symbolic logic than there is for Sudoku, but you can probably do just fine in the world without either.

With symbolic logic, we're really only looking at validity and consistency. In fact, with a formal proof (like the example I gave above) it is not possible to prove invalidity. If it turns out the premises do not warrant the conclusion, then attempting a formal proof will only incline us to suggest that it isn't valid. In order to prove invalidity, we use short hand truth tables. Basically, we write out each premise and the conclusion, then we attempt to find a way in which it is possible for the premises to be true and its conclusion false by applying true or false values to its simple components.

E.g.:

1) A & B
2) ~B --> C
Conclusion: C

Then we'd write each component--namely, A, B and C--and apply T (true) or F (false) to each of them to see if we can show invalidity.

Once again, this is an overly simple example. But let's say we apply T, T and F to A, B and C, respectively. For premise 1, A is true and B is true, and since it is an "and" statement (wherein both parts must be true in order for the whole statement to be true) we then have premise 1 as true. In premise 2, the antecedent is not-B (~B). So, if B is true (as we designated it) then not-B is false. The consequent of the conditional in premise 2 is C, which we designated as false. The only way a conditional statement can turn out false is if the antecedent is true and the consequent is false. Any other combination deems the whole statement true. Therefore, by applying true to B and false to C, premise 2 comes out true. Finally, since we've applied false to C, the conclusion is false. Therefore, we have shown how the premises can be true while the conclusion is false, thus proving invalidity.

Short hand truth tables can also prove consistency. That is, they can prove that a set of statements can consistently come out true by applying truth values to their components. Formal proofs cannot do this. Whereas short hand truth tables can prove invalidity and consistency, formal proofs can prove validity and inconsistency.

Remember, validity has to do with the connection between the premises and the conclusion. It hasn't to do with whether the statements reflect reality. We only learn how to translate from English into "L" in order to understand what we're looking at. Once we're totally in "L," we're no longer concerned with what English might be applied to the symbols. In other words, think of the translating part as mere scaffolding. Consequently, we're not concerned with the overall soundness of an argument, which involves determining whether the premises reflect reality. We're just concerned with the logical form; the implied inference between the premises and conclusion.

I'll end this here because I have to go. But later on today I'll give another example of a formal proof that is slightly more complex than my previous example.

The above does not prove validity. It proves invalidity. But I understand where you're coming from about it being overcomplicated. It is especially overcomplicated because the example is especially over-simple, as I pointed out. If we were to plug English statements into it, one could easily grasp that it lacks validity. But I've been using simple examples thus far only because I'm just trying to introduce the concepts.

A more complex example of a formal proof is as follows:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)

You might notice, right off the bat, that there are no simple components to work with here. Therefore, we can't work top-down, so to speak. If, for example, we had as a premise, J, then we could derive (S & G) from premise 2. Since we have no simple components to start out with, we have to approach this proof with what's called the bottom-up method. In other words, we have to assume something.

You can make two kinds of assumptions in propositional logic: conditional proof and indirect proof. A conditional proof is where you assume the antecedent to a conditional and try and derive its consequent. An indirect proof is when you assume the opposite of what you're trying to prove to be true. The theological problem of evil argument utilizes indirect proof, also known as Reductio Ad Absurdum. Anselm's ontological argument for the existence of God does, as well. With an indirect proof, we are looking to derive a contradiction of any kind. If, for instance, we assume B and, from this, are able to derive F & ~F, then we are allowed to leave the subproof by writing the negation of that which we assumed. In this case, we'd write ~B. Once a subproof is closed, we are not allowed to use any of those lines in our proof.

So which type of assumption should we use? I'd say our best bet is to use conditional proof. Although we can likely use indirect proof to derive some simple statement with which to begin breaking things apart, since the conclusion is a conditional itself, it is usually best to begin by assuming the antecedent of the conclusion, J, as a conditional proof:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)

Whenever we make an assumption, we indent. Due to formatting issues on this forum, I'm filling in the gap with periods. Also, over on the right side we write "assume" and then, in parentheses, an abbreviation/acronym of the type of assumption we're making. In this case, CP for conditional proof. By asuming J as a conditional proof, we're attempting to derive ~R --> Q. Once we have this, we can leave this subproof. Moreover, once we have this, we have the conclusion and are finished with our proof.

As previously stated, since we're in a subproof, we must remain in it until we derive the consequent of the conditional statement we're trying to prove. In other words, line 5 will also remain indented. Also, whenever we're in a subproof, we are only allowed to utilize lines that are directly above or to the left of the subproof. And although we must continue to write our lines indented, we can utilize all lines (e.g. 1, 2 and 3). Since we now have J to work with, we see that on line 2, we have a conditional statement where the antecedent is J. So, the next move is simple:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP

We derive line 5 from lines 2 and 4 by the rule of modus ponens (MP).

At this point, we have to see how we might be able to use the simple components S and G. We notice that in premise 3 there is an S. However, it is currently stuck in a conjunction, and since we need both parts of the conjunction in order to play with that conditional, we are unable to do anything with premise 3. Once again, we have no top-down moves to make. We must make another assumption; a subproof within the subproof. We might notice that the conclusion we're trying to prove is a conditional statement within a larger conditional statement. We've already assumed the antecedent of the overall conditional on line 4. So the best next step would be to assume the antecedent of the conditional statement that appears in the conclusion after the J, namely, ~R. In other words, the conclusion is J --> (~R --> Q), we've assumed J and gone as far as we can, and so now we can assume ~R as another conditional proof in order to derive Q:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP
6. ...................~R.............................................Assume (CP)

The same rules apply as before: we can utilize all lines of the proof (since they are all directly up or to the left of the subproof) but we can't leave this new subproof until we've derived the conclusion we're looking for, namely, Q.

Once again, we look to see what ~R can do for us, top-down. Premise 1 is the only premise that houses an R. Lucky for us, there is a rule for conditionals known as modus tollens. Whereas modus ponens is used to derive the consequent by use of the antecedent, modus tollens is used to derive the negation of the antecedent by use of the negation of the consequent. An example in English might be as follows:

1. If Jason lives in Los Angeles, then he lives in California.
2. Jason does not live in California.
Therefore: Jason does not live in Los Angeles.

Beware of the fallacies known as asserting the consequent and denying the antecedent. Asserting the consequent looks as follows:

1. If Jason lives in Los Angeles, then he lives in California.
2. Jason lives in California.
Therefore: Jason lives in Los Angeles.

Obviously this does not follow and can be understood via the English since we know that other cities exist within California. Jason might, for instance, live in San Francisco.

The fallacy of denying the antecedent goes:

1. If Jason lives in Los Angeles, then he lives in California.
2. Jason does not live in Los Angeles.
Therefore: Jason does not live in California.

This does not follow for pretty much the same reason as above. Jason's not living in Los Angeles, CA does not, therefore, bar him from living in any other city within the state of California.

Back to our formal proof:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP
6. ...................~R.............................................Assume (CP)
7. ...................~H.............................................1, 6 MT

Again, line 7 comes from lines 1 and 6 by the rule of modus tollens. Now we look to see where we can utilize ~H. Premise 3 has it within the conjunction as the antecedent of a conditional statement. We realize that in order to use modus ponens on premise 3, we need both ~H and S. And, in fact, we already have both. The next step is to extract the S from line 5 by the rule of simplification so that we may utilize it on its own:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP
6. ...................~R.............................................Assume (CP)
7. ...................~H.............................................1, 6 MT
8. ...................S...............................................5 Simp

Line 8 comes from line 5 by the rule of simplification (Simp). So now we have the ~H and the S by themselves. We want them together as a conjunction in order to play with the conditional in line 3. This move is also very simple. Just as we can take off parts of a conjunction at will, we can put statements together as conjunctions at will. Of course, we could write it as S & ~H, but we want the conjunction to look as it does in premise 3. So the next step looks as follows:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP
6. ...................~R.............................................Assume (CP)
7. ...................~H.............................................1, 6 MT
8. ...................S...............................................5 Simp
9. ...................~H & S.......................................7, 8 Conj

Line 9 comes from lines 7 and 8 by the rule of conjunction (Conj). The next step we've already discussed:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP
6. ...................~R.............................................Assume (CP)
7. ...................~H.............................................1, 6 MT
8. ...................S...............................................5 Simp
9. ...................~H & S.......................................7, 8 Conj
10. .................Q...............................................3, 9 MP

Line 10 comes from lines 3 and 9 by the rule of modus ponens. If you'll remember, the current subproof assumed the antecedent ~R in order to try and derive the consequent Q. We have now accomplished that task. Therefore, we are now allowed to exit this subproof. However, we are still within the first subproof. When we leave the subproof of a conditional proof, the very next line must be the conditional we were trying to prove. In this case, ~R --> Q:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP
6. ...................~R.............................................Assume (CP)
7. ...................~H.............................................1, 6 MT
8. ...................S...............................................5 Simp
9. ...................~H & S.......................................7, 8 Conj
10. .................Q...............................................3, 9 MP
11. .......~R --> Q...............................................6, 10 CP

Line 11 comes from lines 6 and 10 by conditional proof. Since we have left the subproof that began on line 6 and ended on line 10, we are no longer able to utilize lines 6 through 10. They are now off limits. In other words, in no future lines can we refer back to 6, 7, 8, 9 or 10. That subproof is closed.

However, we are still within the subproof started on line 4. Looking back at line 4, we realize that we were assuming J as the antecedent of a conditional proof where the consequent being pursued was ~R --> Q. Well, we have that now on line 11. So, just as fast as we left our right-most subproof, we can leave this one:

1. H --> R
2. J --> (S & G)
3. (~H & S) --> Q................................................Therefore: J --> (~R --> Q)
4. .........J..........................................................Assume (CP)
5. .........S & G....................................................2, 4 MP
6. ...................~R.............................................Assume (CP)
7. ...................~H.............................................1, 6 MT
8. ...................S...............................................5 Simp
9. ...................~H & S.......................................7, 8 Conj
10. .................Q...............................................3, 9 MP
11. .......~R --> Q...............................................6, 10 CP
12. J --> (~R --> Q)............................................4, 11 CP

Line 12 comes from lines 4 and 11 by conditional proof. And we're finished. We've proven that, given the premises, we can derive the conclusion. In other words, we've proven validity.

When did I ever say I was a creationist?