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Here is an example of a De Morgan transformation of a logic expression:

$ \neg ({a} \wedge {b}) $

becomes

$ (\neg {a} \vee \neg {b}) $

My intuitive view of this operation is that I'm "moving the negation" into the parenthesis and changing the operator into its duel ( $ \wedge $ to $ \vee$ ).

However, I'm curious about how operations like this would be applied to expressions that are expressed in infix, without parenthesis. I can tell that the example I gave above would look something like this:

$ a \space b \wedge \neg $

becomes

$ a \space \neg \space b \space \neg\vee $

Yet, it isn't obvious to me how to systematically figure out where in the stack the negation operators should go. I could probably figure it out and feel good about the answer, but I don't know how to program a computer to do it.

I could easily come up with more complicated examples by making a and b expressions of more variables and by adding more than just two sets of expressions. Like this: $ (a_1 \wedge a_2 \wedge a_3) \vee ( b_1 \wedge b_2 \wedge b_3 ) \vee (c_1 \wedge c_2 \wedge c_3) $

or in infix form: $ a_1 \space a_2 \wedge a_3 \wedge b_1 \space b_2 \wedge b_3 \wedge c_1 \space c_2 \wedge c_3 \wedge \vee \vee $

So, what would be a systematic way of figuring out how to transform the expression? Or at least, a way to understand that process better?

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    $\begingroup$ Parse the expression into a tree form, and then the negation moves between some node and its children. $\endgroup$ – dtldarek Dec 8 '14 at 23:13
  • $\begingroup$ Parsing it into a tree does make sense. $\endgroup$ – beeflobill Dec 8 '14 at 23:35
  • $\begingroup$ But, how about in terms of pencil and paper? Say, if we grew up in an alternate universe and we all learned polish-notation instead of ... non-polish-notation. $\endgroup$ – beeflobill Dec 8 '14 at 23:51
  • $\begingroup$ (you (((no problems) (the expression) (parsing subconsciously)) (you now ((notation infix) with) do) as)) (have would) Then. $\endgroup$ – dtldarek Dec 9 '14 at 0:31
  • $\begingroup$ The message I'm getting is that you think I'm a simplistic weenie and I'm wasting time on something worthless, or at least, ill defined. I think this is mostly the fault of how I asked the question. $\endgroup$ – beeflobill Dec 9 '14 at 16:04
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First, you've compared expressions in reverse Polish notation and an infix notation here.

Second, under some interpretations a1 a2∧a3∧b1 b2∧b3∧c1 c2∧c3∧∨∨ is more precise than (a1∧a2∧a3)∨(b1∧b2∧b3)∨(c1∧c2∧c3), since the order of parsing all strings in (a1∧a2∧a3)∨(b1∧b2∧b3)∨(c1∧c2∧c3) is not clear (do we parse a1∧a2 first or a2∧a3?). At the same time that parsing order isn't relevant to say computing (a1∧a2∧a3)∨(b1∧b2∧b3)∨(c1∧c2∧c3) since both (S, $\land$) and (S, $\lor$) are commutative semigroups.

Third, in Reverse Polish notation for any expression where the main function is binary, we have a left argument and a right argument. Let's look at an algorithm to determine whether or not an expression is in Reverse Polish notation.

  1. For any n-ary function or predicate symbol assign (n-1) to that symbol.
  2. Assign -1 to all variables.

Thus, $\lor$ gets assigned "1", and $\lnot$ gets assigned 0.

Now start at the end of the expression and assume we have 0 before we start counting. An expression is a well-formed expression in Reverse Polish notation if and only if it is a variable, or starts with a natural number on the right and ends with -1 on the left and only gets to -1 at the last position on the left.

Now if we have a binary function as the symbol on the far right of some well-formed expression, then we'll have a left expression and a right expression. It turns out that the right expression has a "0" corresponding to the symbol in the leftmost position, and the left expression has a "-1" corresponding to the symbol in the leftmost position. Determine the right expression first, and then the left expression second For example, let A be binary, K binary, and N unary... the assignments above for xyAzNuNNKA give us the following:

x   y   A  z  N  u  N  N  K  A
|   |   |  |  |  |  |  |  |  |
-1  0   1  0  1  1  2  2  2  1

Thus, zNuNNK is the right expression of the main "A" expression, and xyA is the left expression of the main "A" expression. Thus, since xyKN=xNyNA and xyAN=xNyNK, if we wanted to find say how to figure out where the negation in

xyzKAuvKwzyKAAKN goes we just figure out what the left and right expressions of xyzKAuvKwzyKAAK are. Then we put an "N" after each of those expressions, and concatenate them accordingly (corresponding to how the non-negated left expression and right expression were concatenated) and switch the main function to its dual.

For an expression with just binary functions like ((a1∧(a2∧a3))∨(((b1∧b2)∧b3)∨((c1∧c2)∧c3))), we look at how it gets constructed from its sub-expressions. At the first level we have (a2∧a3), (b1∧b2), and (c1∧c2). At the second level we have (a1∧(a2∧a3), ((b1∧b2)∧b3), and ((c1∧c2)∧c3). At the third level we have (((b1∧b2)∧b3)∨((c1∧c2)∧c3)). And at the fourth level we have ((a1∧(a2∧a3))∨(((b1∧b2)∧b3)∨((c1∧c2)∧c3))). So, we can translate each expression at level 1 from (x?y) to xy? where "?" is some binary function. Then we translate all level 2 expressions, and so on, until no more levels are left. But, let's call $\land$, K when in Reverse Polish notation, and $\lor$, A so we don't get confused when translating. With this example that would mean we start with

((a1∧(a2∧a3))∨(((b1∧b2)∧b3)∨((c1∧c2)∧c3))) which becomes

((a1∧a2a3K)∨((b1b2K∧b3)∨(c1c2K∧c3)))

after transforming all level 1 expressions. Now transforming the next level we obtain

(a1a2a3KK∨(b1b2Kb3K∨c1c2Kc3K))

Transforming the next level we obtain

(a1a2a3KK∨b1b2Kb3Kc1c2Kc3KA)

And thus finally

a1a2a3KKb1b2Kb3Kc1c2Kc3KAA.

Parsing the expression into a tree should help to determine the level of a sub-expression.

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