Let there be a composite statement R, consisting of 2 component statements P and Q. There are 4 possibilitues for P and Q, they are either (T,T), (T,F), (F,T), or (F,F). Or let T=1 and F=0, and replacing the variables you'd have (1,1), (1,0), (0,1), or (0,0). Now depending on the composite statement R, each possibility for PxQ will lead to either 1 or 0 for R, so each composite statement R should have its own individual set of 4 values, where it's mapping each element of PxQ to either 1 or 0. I'm having some trouble describing these values. They are the output values for the elements of PxQ I listed, but when I evaluate these composite statements, I want to evaluate the output in the order I listed the elements of the input. So it's not exactly a set. Could I call it an ordered set of outputs?

The point I'm trying to make is that no matter how complicated or large a composite statement you make is that contains exactly 2 component statements, it's logically equivalent to exactly one of these 16 composite statements enter link description here

I differentiate these composite statements (or really functions), by these 'ordered sets'. The elements in a set can be arranged however one wants and not have an effect on the set, however the placement of the elements in these is attempting to show which element of PxQ they are related to.

Also I think it's interesting that if you look at any of these types of logical connectives( implication, disjunction, conjuction, negation) you can disregard their philosophical or language meaning and considered them to just be a function that maps the elements in PxQ to a specific ordered set. Combining the various logical connectives can allow you to create different functions. But if I picked or made 3 or 4 different logical connectives, which were some of these other functions, could I create the rest of the 16 functions that are logically equivalent to all composite statements that have exactly 2 component statements?

Also it seems that composite statements with n components have 2^n elements in the input and output, and that there are 2^(2^(n)) composite statements which all composite statements with n component statements are logically equivalent to exactly one of?

  • $\begingroup$ Maybe I have misunderstood the question, but it seems to me that in the third paragraph you yourself have given the word you need: "function". $\endgroup$ – David Aug 15 '16 at 4:22
  • $\begingroup$ I'm trying to describe the 'ordered sets'. I don't know what I should call them, just some word I came up with. I needed quite a bit of lines to describe what I meant, I want to describe them in a way thats far more concise, that potentially describes them better. It's like listing the elements in the domain in a specific order and listing the elements in the range in a specific order that matches with the inputs that map to them. Basically what a truth table does. $\endgroup$ – Hockeyfan19 Aug 15 '16 at 4:28
  • $\begingroup$ I'm also curious if I'm on the right track with the questions I asked. I'm not that knowledgeable concerning math, I'm just trying to understand stuff. $\endgroup$ – Hockeyfan19 Aug 15 '16 at 4:30

As mentioned in a comment, what you are describing are functions, specifically truth functions, which in classical logic are functions from a given number of boolean inputs to a single boolean output. If there are $n$ inputs, there are clearly $2^{2^n}$ possible truth functions. Note that it is possible to obtain all possible truth functions using any functionally complete set of boolean connectives.

| cite | improve this answer | |
  • $\begingroup$ Thanks, that Wikipedia link answered nearly all of my questions. Actually looking at it it seems that there are 2^(2^n) truth functions in n-valued logic, where n is the number of component statements. For example if you have 3 component statements, you'd have 256 truth functions, or in two-valued logic you have 16 like in the link. Or "Boolean Functions". I've heard that word a lot, but I'm happy to finally understand what it means. Thanks a lot! $\endgroup$ – Hockeyfan19 Aug 15 '16 at 9:23
  • $\begingroup$ @Hockeyfan19: You're right; I was so careless; it's definitely $2^{2^n}$, since there are $2^n$ possible input combinations. Haha! $\endgroup$ – user21820 Aug 15 '16 at 9:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.