friend wrote:OK, I have a few misstatements that I have to correct. I wrote " If the conjunctions (^) can be replaced with summations and the disjuctions (V) replaced by multiplications". This should be, " If the conjunctions (^) can be replaced with multiplications and the disjuctions (V) replaced by summations". Sorry about that. It seems I read what I meant to write and not what I actually wrote. I only catch these mistakes if I let enough time pass before rereading it so I can forget what I intended to write, which is difficult. Here is a corrected version.

What does physics have to do with logic? Physics has to do with what exists or not. Logic has to do with what is true or false. Existence and nonexistence are mutually exclusive binary states just as true and false are mutually exclusive binary states. So logic can be used to symbolically represent what exists or not. This is actually quite obvious. For we use propositions, which are either true or false, to describe physical situations which may exist or not. This is simply the correspondence principle used as the definition of truth, that reality is described by propositions which are true. This is the starting premise of my work. It's really quite simple, even advanced high school students can follow along.

I start by recognizing that each point in space is different from every other point in space. We recognize this when we assign different coordinates to each point. But this is actually a logical relationship. For what is being said is that this point here is part of space AND this other point here is part of space AND this still other point is part of space AND... This is expressed as a conjunction (^) of all points as shown next.

,

where each

q represents its own point in space. Obviously there is an infinite number of them.

It is then easy to show that a conjunction (^) implies a material implication (

) as shown next

.

My website (logictophysics.com) points to a place on the web where you can enter logical expressions and have a truth-table generated for it. This makes it easy to check my work.

And if we apply this to the entire conjunction, we get,

.

But I also show on my website that,

.

I slow walk my readers to this equation with simple expressions they can plug into a truth-table generator. All it requires is a little patience. But here you can start to see the beginnings of the path integral. If the conjunctions (^) can be replaced with multiplications and the disjuctions (V) replaced by summations and the implications

replaced by transition amplitudes, then the path integral will result. The question is how to do this.

One reason why you may not understand what Boolean logic has to do with physics is because you don't understand how numbers can be introduced to logic. Logic and numbers in your mind are completely divorced from each other. So physics, which is mathematical, cannot possibly be connected with symbolic logic. I remedy this by using the Dirac measure to get the Kronecker delta. The Dirac measure assigns a value of 1 if the element is included in the set A, and is 0 otherwise.

Set inclusion represents implication very well. For if the entire set is said to exist, then so can its element be said to exist. But just because an element may exist does not mean that the entire set exists. So we can replace implication in the logic with the Dirac measure. And this brings in numbers into the logic.

But in the last logic expression above, a proposition used as a premise in one implication can just as easily be used as a conclusion in some other implication. So let the set, A, of the Dirac measure be shrunk down to near the size of an element so that the set A can be represented by a number as well. Then we get the Kronecker delta,

Now that we have numbers, how do we get the math of addition and multiplication? As I explain on my website, a proposition,

q, can be written as Pq, which is true if the object q has the property P and is false if q does not have the property P. But the property, P, has the

extension P={q1,q2,q3,...}, the set of all such objects that have that property. This means that

q is true if (q ∈ P) and

q is false if (q ∉ P). When I use set inclusion to represent implication, (

P q) is equal to (q ∈ P) for which the Dirac measure is

. In terms of the extension of P={q1,q2,q3,...qn}, then (

P qj) =

T only if 1

jn, and is

F if

j1 or

jn. If P has only one element, qk, then the implication,

qjqk=

T only for

j=k and is

F otherwise. Then we have the following:

.

If

j is either 1 or 2, then the expression is true. If

j is outside this range, then it is false. This is because for either

j=1 or

j=2, the expression reduces to (q1

q1)V(q2

q1) = T V F = T, or it reduces to (q1

q2)V(q2

q2) = F V T = T. But if

j is outside this range, then both (q1

qj) and (q2

qj) are false. So if the analogy with the Kronecker delta is to be upheld, then we are looking for a math expression for disjunction (let's call it

for now) such that the following expression results. Remember, we can replace

T for 1 since

T = (qx

qx) which is replace by

= 1.

When we replace the implications with Kronecker deltas and the

T with 1 and the

F with 0, the following table results for the various conditions for

j.

Notice that there is never the case where

j = 1 and

j = 2 at the same time. So we will never have the case where

= 1 and

= 1 at the same time. It is obvious from this table that

should be represented by the math operator of addition +.

How do we find the math operator for conjunction? Consider the logic expression below:

.

If we replace the implications with Kronecker deltas and the disjunctions with addition, then we get,

,

with,

, as some yet unknown math operator for conjunction. With

n=1 in the last expression,

k can only have the value of 1, and the last expression reduces to,

. With

i or

j any value other than 1, this expression is 0. And we get the following table for the specified conditions:

.

Obviously,

represents the math operator of multiplication. And putting it all together, the logic expression,

,

becomes the math expression,

.

I think it is a typical step to go from the discrete case with the Kronecker delta to the continuous case with the Dirac delta function. From there it's a matter of finding reasons why to select the complex exponential Gaussian to represent the Dirac delta function. But when this is done, the Feynman Path Integral results. There is some debate as to why the complex exponential Gaussian function should be chosen for the Dirac delta function. I have shown on my website some obvious reasons for selecting it.

I include a more detail on my website at: logictophysics.com

[quote="friend"]OK, I have a few misstatements that I have to correct. I wrote " If the conjunctions (^) can be replaced with summations and the disjuctions (V) replaced by multiplications". This should be, " If the conjunctions (^) can be replaced with multiplications and the disjuctions (V) replaced by summations". Sorry about that. It seems I read what I meant to write and not what I actually wrote. I only catch these mistakes if I let enough time pass before rereading it so I can forget what I intended to write, which is difficult. Here is a corrected version. [/quote]

What does physics have to do with logic? Physics has to do with what exists or not. Logic has to do with what is true or false. Existence and nonexistence are mutually exclusive binary states just as true and false are mutually exclusive binary states. So logic can be used to symbolically represent what exists or not. This is actually quite obvious. For we use propositions, which are either true or false, to describe physical situations which may exist or not. This is simply the correspondence principle used as the definition of truth, that reality is described by propositions which are true. This is the starting premise of my work. It's really quite simple, even advanced high school students can follow along.

I start by recognizing that each point in space is different from every other point in space. We recognize this when we assign different coordinates to each point. But this is actually a logical relationship. For what is being said is that this point here is part of space AND this other point here is part of space AND this still other point is part of space AND... This is expressed as a conjunction (^) of all points as shown next.

[img]http://logictophysics.com/conjqsb.gif[/img],

where each [i]q[/i] represents its own point in space. Obviously there is an infinite number of them.

It is then easy to show that a conjunction (^) implies a material implication ([img]http://logictophysics.com/matimpl.gif[/img]) as shown next

[img]http://logictophysics.com/ceqim.gif[/img].

My website (logictophysics.com) points to a place on the web where you can enter logical expressions and have a truth-table generated for it. This makes it easy to check my work.

And if we apply this to the entire conjunction, we get,

[img]http://logictophysics.com/allcnimimp.gif[/img].

But I also show on my website that,

[img]http://logictophysics.com/totimeq.gif[/img].

I slow walk my readers to this equation with simple expressions they can plug into a truth-table generator. All it requires is a little patience. But here you can start to see the beginnings of the path integral. If the conjunctions (^) can be replaced with multiplications and the disjuctions (V) replaced by summations and the implications [img]http://logictophysics.com/implication.gif[/img] replaced by transition amplitudes, then the path integral will result. The question is how to do this.

One reason why you may not understand what Boolean logic has to do with physics is because you don't understand how numbers can be introduced to logic. Logic and numbers in your mind are completely divorced from each other. So physics, which is mathematical, cannot possibly be connected with symbolic logic. I remedy this by using the Dirac measure to get the Kronecker delta. The Dirac measure assigns a value of 1 if the element is included in the set A, and is 0 otherwise.

[img]http://logictophysics.com/Dmeas.gif[/img]

Set inclusion represents implication very well. For if the entire set is said to exist, then so can its element be said to exist. But just because an element may exist does not mean that the entire set exists. So we can replace implication in the logic with the Dirac measure. And this brings in numbers into the logic.

But in the last logic expression above, a proposition used as a premise in one implication can just as easily be used as a conclusion in some other implication. So let the set, A, of the Dirac measure be shrunk down to near the size of an element so that the set A can be represented by a number as well. Then we get the Kronecker delta,

[img]http://logictophysics.com/KronDelt.gif[/img]

Now that we have numbers, how do we get the math of addition and multiplication? As I explain on my website, a proposition, [b][i]q[/i][/b], can be written as Pq, which is true if the object q has the property P and is false if q does not have the property P. But the property, P, has the [i]extension[/i] P={q1,q2,q3,...}, the set of all such objects that have that property. This means that [b][i]q[/i][/b] is true if (q ∈ P) and [i][b]q[/b][/i] is false if (q ∉ P). When I use set inclusion to represent implication, ([b][i]P[/i][/b] [img]http://logictophysics.com/matimpl.gif[/img] [b][i]q[/i][/b]) is equal to (q ∈ P) for which the Dirac measure is [img]http://logictophysics.com/DMqP.gif[/img]. In terms of the extension of P={q1,q2,q3,...qn}, then ([i][b]P[/b][/i] [img]http://logictophysics.com/matimpl.gif[/img] [i][b]q[/b]j[/i]) = [b][i]T[/i][/b] only if 1[tex]\le[/tex][i]j[/i][tex]\le[/tex][i]n[/i], and is [i][b]F[/b][/i] if [i]j[/i][tex]<[/tex]1 or [i]j[/i][tex]>[/tex][i]n[/i]. If P has only one element, qk, then the implication, [i][b]q[/b]j[/i][img]http://logictophysics.com/matimpl.gif[/img][i][b]q[/b]k[/i]=[i][b]T[/b][/i] only for [i]j=k[/i] and is [i][b]F[/b][/i] otherwise. Then we have the following:

[img]http://logictophysics.com/Orisadd1.gif[/img].

If [i]j[/i] is either 1 or 2, then the expression is true. If [i]j[/i] is outside this range, then it is false. This is because for either [i]j[/i]=1 or [i]j[/i]=2, the expression reduces to (q1[img]http://logictophysics.com/matimpl.gif[/img]q1)V(q2[img]http://logictophysics.com/matimpl.gif[/img]q1) = T V F = T, or it reduces to (q1[img]http://logictophysics.com/matimpl.gif[/img]q2)V(q2[img]http://logictophysics.com/matimpl.gif[/img]q2) = F V T = T. But if [i]j[/i] is outside this range, then both (q1[img]http://logictophysics.com/matimpl.gif[/img]qj) and (q2[img]http://logictophysics.com/matimpl.gif[/img]qj) are false. So if the analogy with the Kronecker delta is to be upheld, then we are looking for a math expression for disjunction (let's call it [img]http://logictophysics.com/adddd.gif[/img] for now) such that the following expression results. Remember, we can replace [b]T[/b] for 1 since [b]T[/b] = (qx[img]http://logictophysics.com/matimpl.gif[/img]qx) which is replace by [tex]{\delta _{xx}}[/tex] = 1.

[img]http://logictophysics.com/Orisadd2.gif[/img]

When we replace the implications with Kronecker deltas and the [b]T[/b] with 1 and the [b]F[/b] with 0, the following table results for the various conditions for [i]j[/i].

[img]http://logictophysics.com/Orisadd3.gif[/img]

Notice that there is never the case where [i]j[/i] = 1 and [i]j[/i] = 2 at the same time. So we will never have the case where [tex]{\delta _{11}}[/tex] = 1 and [tex]{\delta _{22}}[/tex] = 1 at the same time. It is obvious from this table that [img]http://logictophysics.com/adddd.gif[/img] should be represented by the math operator of addition +.

How do we find the math operator for conjunction? Consider the logic expression below:

[img]http://logictophysics.com/conmult1.gif[/img].

If we replace the implications with Kronecker deltas and the disjunctions with addition, then we get,

[img]http://logictophysics.com/conmult2.gif[/img],

with, [img]http://logictophysics.com/conmult3.gif[/img], as some yet unknown math operator for conjunction. With [i]n[/i]=1 in the last expression, [i]k[/i] can only have the value of 1, and the last expression reduces to,

[img]http://logictophysics.com/conmult4.gif[/img]. With [i]i[/i] or [i]j[/i] any value other than 1, this expression is 0. And we get the following table for the specified conditions:

[img]http://logictophysics.com/conmult5.gif[/img].

Obviously, [img]http://logictophysics.com/conmult3.gif[/img] represents the math operator of multiplication. And putting it all together, the logic expression,

[img]http://logictophysics.com/logicpath.gif[/img],

becomes the math expression,

[img]http://logictophysics.com/deltapath.gif[/img].

I think it is a typical step to go from the discrete case with the Kronecker delta to the continuous case with the Dirac delta function. From there it's a matter of finding reasons why to select the complex exponential Gaussian to represent the Dirac delta function. But when this is done, the Feynman Path Integral results. There is some debate as to why the complex exponential Gaussian function should be chosen for the Dirac delta function. I have shown on my website some obvious reasons for selecting it.

I include a more detail on my website at: logictophysics.com