Logical Consistency of Belief (or the Lack Thereof)

Last post, I showed that it is possible to believe two contradictory statements at the same time. Now I'm going to go a step farther and show that none of the rules of logic necessarily apply to beliefs at all.


First, an informal, verbal argument; there's a more rigorous proof in prop calc below. We discussed before the difference between believing a statement is false, and not believing the statement. The fundamental difference can be expressed thusly: When you believe that a statement is false, you are in a state of belief -- namely, a state of believing the statement's negation. When you don't believe the statement, you are not in a state of belief.

It is possible to believe two contradictory statements, but impossible to believe and not believe the same statement. This is because in the former case, you're in a state of believing the first statement, and in a state of believing the second statement. There's nothing inherently contradictory about being in a state of believing one statement and simultaneously being in a state of believing another statement. However, to believe and not believe the same statement would require you to be in a state, and also not be in the same state, which is inherently contradictory.

So, let's assume that the laws of logic apply to beliefs: that if you believe a statement, you also believe its logical consequences. (Obviously, since we've already established that it's possible to believe contradictory statements, the law of non-contradiction is excepted.) Well, now we've got a problem, because through a simple proof with the delightful name "Principle of Explosion," you can show that any and every statement follows from a logical contradiction. In other words, if beliefs always behave logically, then anyone who believes two contradictory statements must therefore believe everything.

Now, none of us has ever met a person who literally believes everything. But maybe that just proves that, while it's logically possible to believe two contradictory statements, for some reason nobody ever does? Except for one problem: plenty of people do believe contradictory statements, in the form of religious mysteries. "God is three" and "God is one", for example. "God is all-powerful, all-knowing, and all-loving," and "Suffering exists." The most common argument given by believers (not theologians, but believers "on the ground") is that logic does not apply to God. However, in order to believe in a being that transcends the rules of logic, ones beliefs must also transcend the rules of logic!

In short, over the last two posts we've established that beliefs need not be logically consistent with one another. That will be important in the next post, which I hope to get up this weekend -- and I guarantee that one will be entirely prop-calc free!

Rigorous proof:

The first step is to translate into symbols the statement "Beliefs obey the rules of logic." Now, all rules of inference ultimately take the form p -> q -- if one statement is true, then another statement must also be true. For example, the rule Disjunctive Syllogism can be stated as

((a V b) * ~a) -> b

((a V b) * ~a) is the p, and b is the q.

Any particular logical proof is going to consist a series of steps that take the form:

([Premise(s)] * [Rule of Inference]) -> [Conclusion]

Note that we're referring to the premises and conclusion of the step here, not the premises and conclusion of the entire proof. Using the generic p -> q form of the rules of inference, we can see that one way of stating "the rules of logic apply" is:

(p * (p -> q)) -> q

Unfortunately, we can't quite use this for beliefs. The reason is that the above assumes that the syllogism

Premise: p
Premise: q
Conclusion: p * q

holds. It does for logic, but does it necessarily hold for belief? We honestly have no idea, at this stage, whether (B(p) * B(q)) = B(p * q). Hence, we can't use (B(p) * (p -> q)) * B(q) as a way of saying "logic applies to belief" because it doesn't cover rules of inference with multiple premises, like Modus Ponens and Disjunctive Syllogism.

For those rules of inference, we need to use the form (B(p) * B(q) * ((p * q) -> r)) -> B(r). As we'll see below, this actually covers all the rules of inference. We'll begin with two premises. These create our "test believer" -- we're imagining a person who believes an arbitrary statement p as well as its negation ~p, and who also does not believe an arbitrary statement q. (Remember, although we proved B(p) * B(~p) is not a contradiction, B(p) * ~ B(p) still is one!)

Absurdly, we will have to take a break partway through to demonstrate that (p * q) -> (p * q). For some reason, p -> p is not one of the variants of the rule of inference Tautology. I've never understood why. And since (p * q) -> (p * q) both requires fewer steps to prove, and is the statement we actually need later in the proof, I took the detour.

1. B(p) * B(~p) , Premise (from prior proof)
2. ~B(q), Premise
3. (B(p) * B(q) * ((p * q) -> r)) -> B(r), Assumption
4. B(p) * (p -> r), Assumption
5. p -> r, 4, Simplification
6. (p*p) -> r, 5, Tautology
7. B(p), 4, Simplification
8. B(p) * B(p), 7, Tautology
9. B(p) * B(p) * ((p*p) -> r), 6, 8, Conjunction
10. B(r), 3, 9, Modus Ponens
11. (B(p) * (p -> r)) -> B(r), 4, 10, Induction
12. p * q, Assumption
13. q * p, 12, Commutation
14. p * q, 13, Commutation
15. (p * q) -> (p * q), 12, 14, Induction
16. B(p) * B(~p) * ((p * q) -> (p * q), 1, 15, Conjunction
17. B(p * ~p), 3, 16, Modus Ponens
18. p * ~p, Assumption
19. p, 18, Simplification
20. p V q, 19, Addition
21. ~p, 18, Simplification
22. q, 20, 21, Disjunctive Syllogism
23. (p * ~p) -> q, 18, 22, Induction
24. B(p * ~p) * ((p * ~p) -> q) , 17, 23, Conjunction
25. B(q), 11, 24, Modus Ponens
26. ~B(q) * B(q), 2, 25, Conjunction
27. ~((B(p) * B(q) * ((p * q) -> r)) -> B(r)), 3, 26, Disproof by Contradiction