On page 18 "Logic as Algebra" Halmos&Givant wrote the distributive law in Polish notation as $$ = \times a + bc + \times ab \times ac $$ I fail to see anything remarkable here, is there a combinatorial pattern that I'm missing?
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5What exactly do me mean when you say 'remarkable'? Are you asking how to interpret Polish notation? Or the reason why one might represent things in this way? – matt Nov 22 '11 at 03:58

1After rereading the passage it seems that I should lower my expectations. The authors indicated that they can write distributive law without parenthesis, and that's probably all to it. On a related note, is there any algebraic identity which is syntactically much clearer in PN? – Tegiri Nenashi Nov 22 '11 at 04:39

I think you are right, the main advantage of Polish notation is omission of parentheses. Some formulas in symbolic logic become impenetrable masses of parentheses, so there was a need to do something about it if mere humans were to read it. Another attempt was a system of dots to replace parentheses. – GEdgar Nov 22 '11 at 16:15

I've had a Hewlett Packard 48G (graphing calculator) for years. It uses a stack (and thus reverse polish notation). I have to say it wasn't easy to use at first, but there are some tasks it really speeds up and makes incredibly efficient. However, given we're "all" trained to regularly use infix notation, mental gymnastics are needed to more over into the world of prefix/postfix notation. And programming in the world of prefix or postfix notation, ouch, my head hurts. :( – Bill Cook Nov 22 '11 at 19:54

1Programming field suggested several insights how to handle expression complexity, and PN is not one of them. Nested expressions are just parse trees which structure can be emphasized with generous use of carriage return and proper identation. – Tegiri Nenashi Nov 23 '11 at 01:11
3 Answers
I'm not quite sure what you're looking for, but here's a bit about the distributive laws.
The polish notation: $=\times a+bc+\times ab \times ac$ in standard infix notation is: $a \times (b+c) = a \times b + a \times c$.
What this means is one gets the same result if one multiplies then adds or adds then multiplies.
Using abstract algebra terminology, the distributive laws say that multiplication operators are group homomorphisms addition preserving maps. An group homomorphism (using additive notation) addition preserving map is a function such that $\varphi(b+c)=\varphi(b)+\varphi(c)$ (you can add then map or map then add). So in polish notation this is: $= \varphi + b \; c + \varphi \; b \; \varphi \; c$. This is the general pattern you're looking for (I guess).
This property is capturing a kind of commutativity: It doesn't matter which is done first: map or add. Or in your context, it doesn't matter which is done first: add or multiply.
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Helpful, thank you: http://mathoverflow.net/questions/62567/geometricinterpretationofdistributivity – Tegiri Nenashi Nov 22 '11 at 04:57

Do you perhaps mean that the multiplication operators satisfy the homomorphic equation? Specifically, I don't see why you've said "group homomorphisms", since we could have another structure here. – Doug Spoonwood Nov 22 '11 at 19:37

Also, I don't see how the distributive laws correspond to homomorphisms. If we have F(a+b)=F(a)+F(b), then F is unary, and + is binary. In the distrubitve laws, both operations are binary. Note that no matter which notation you pick, the homomorphic equation doesn't match that of the distributive law also, so long as the arity of all symbols comes as clear. – Doug Spoonwood Nov 22 '11 at 19:45

@DougSpoonwood you're right. When I wronte my answer I was thinking of operating in the context of a ring. I guess I should use a more general term like "$\varphi$ is an addition preserving map." – Bill Cook Nov 22 '11 at 19:46

3After *fixing* an element, say $a$, the left multiplication operator $L_a(x)=ax$ is unary and $L_a(x+y)=L_a(x)+L_a(y)$ so the left multiplication operator (for a fixed element) is a group homomorphism. – Bill Cook Nov 22 '11 at 19:48
The Wikipedia article Polish notation describes how Polish notation works. It's not that hard to relearn how to write things in Polish notation and interpret what expressions written in Polish notation mean and use them to derive other results and write them in Polish notation. In fact, after you spend enough time doing things the new way, you'll probably actually find that the Polish notation is more straight forward than the Infix notation.
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You are answering a 7.5 year old question, and are doing so by posting a link to Wikipedia and posting your opinion about what notation is good or bad; even if the question were new, it would not be an appropriate answer. For a question with no activity for so long... – Arturo Magidin Jun 23 '19 at 04:03

@ArturoMagidin I couldn't be absolutely sure a new answer to an old question would never be a useful answer. I think it's fine to try things and learn. I wrote an answer to https://math.stackexchange.com/questions/1820552/wellorderingsofmathbbrwithoutchoice at a time when the question was about 2 years old. At first, AsafKaragila wrote a comment saying something like it doesn't add anything 2 years after Noah Schweber's answer then later it got an upvote. That comment later got deleted. People can't figure out everything with certainty. Your advice might not turn out correct but it's – Timothy Jun 23 '19 at 04:14

fine that you gave it. It's better than no advice. I could end up thinking you're probably right. I'm not sure this answer isn't a good answer that adds something. I'm not sure it is one either. – Timothy Jun 23 '19 at 04:16

I did not say “a new answer to an old question [is] never a useful answer.” **Your** answer here, though, is not particularly useful: you aren’t answering the question asked, instead you used it as a platform for expressing your opinion on notation. So... old question to which you add your unsolicited opinion. Well done. I only regret that I have but one downvote to give. – Arturo Magidin Jun 23 '19 at 05:30

@ArturoMagidin I think the answer at https://math.stackexchange.com/questions/4202/inductiononrealnumbers/4204#4204 tried to figure out what the Author meant and it ended up with being accepted, having a score of 114, and having a bounty of 50 points. I tried to do the same thing here. I think that's the way research works. People don't know very much so they have to guess and say things that might be true and aren't always going to be true. You gave me useful advice that this answer does not answer the question. I'm not sure that's actually the case. Maybe I could have asked the OP what – Timothy Jun 23 '19 at 05:44


And since I'm commenting about the rather poor quality of *this* answer and its context, I find your continued invocation of *other* answers in *other* questions to be an even larger waste of time and space. If that was your objective, well done. Was that answer posted almost 8 years after the question? No. Was that just an expression of the poster's opinion and a "let me Wikipedia that for you" link? No. It is rather disingenuous of you to compare this answer to that. And that is all I will say; if you wish to continue tryiing to justify this waste, go for it. – Arturo Magidin Jun 23 '19 at 05:51

@ArturoMagidin You're right. I didn't figure out that this counted as a let me Google that question for you. I don't know absolutely everything about Polish notation so I'm not sure how to explain it myself in this answer. I'm missing how they deal with problems of this sort. Does +222 mean 2 + 22? Does it mean 22 + 2? I guess I can't think of a useful answer. Do you think I could learn more about Polish notation and explain it properly or is it better to delete this answer. Maybe I could delete it then undelete it if I learn enough about Polish notation. – Timothy Jun 23 '19 at 06:02
I'd say that Polish notation (or reverse Polish notation) makes it clearer that the "the outer operation moves to the inside and doubles, and the inner operation moves to the outside" when applying distributivity. Syntactically speaking, expressions in Polish notation (and reverse Polish notation) come as almost always clearer than in ubiquitous infix notation.
If you consider ax(b+c)=ab+ac, or a(b+c)=(axb)+(axc) and most other infix notation expressions you have to consider more than just the syntax of the expressions involved (such as what the author intends to say, which he may have stated earlier in the text, however, this is not part of the syntax of expressions). In other words, if you interpret either ax(b+c)=ab+ac or a(b+c)=(axb)+(axc) purely in terms of what your formation rules (rules for what wellformed formulas, or formulas) say exactly, they'll come out as nonsensical, or at best as ambiguous. On the other hand in Polish notation, so long as you know the arity of the operations, predicates/relations (for "=") something like =×a+bc+×ab×ac comes as clear from basically the formation rules.
For instance, with this condensed formation rule
 If "p" and "q" all represent welldefined values, then +pq, ×pq, and =pq each represent welldefined values.
and this axiom:
 "a", "b", and "c" represent welldefined values,
then you can formally prove =×a+bc+×ab×ac as welldefined also as follows:
1 +bc by rule 1, and axiom 2
2 ×ab by rule 1, and axiom 2
3 ×ac by rule 1, and axiom 2
4 +×ab×ac by steps 2 and 3, and rule 1
5 ×a+bc by axiom 2, step 1, and rule 1
6 =×a+bc+×ab×ac by steps 3 and 4 and rule 1
You have to fully parenthesize an infix expression in order to do something like this.
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