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I wish to ask some question about natural numbers' construction by using logical trees, without using free variables.
First, some background that leads to my question:
True is notated by 1
~True is notated by 0
p and q are two propositions as follows:
p = 0 0 1 1
q = 0 1 0 1
So, we get the 16 logical connectives as seen by the 16 distinct paths of the following binary tree:
The complements of a given binary tree with 16 distinct paths are:
---------------------------------------------------
Let's briefly touch 3-valued logic.
True has 3 options which are: True, mTrue, ~True (m = middle, ~ = not).
True is notated by 2.
mTrue is notated by 1.
~True is notated by 0.
p, m and q are 3 propositions as follows:
A tree of 3-valued logic of these propositions has 327 = 7,625,597,484,987 logical connectives.
In this case contradiction is path 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0,
where tautology is path 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2.
Moreover, given any n-valued logical tree (where n > 1) it is bounded by contradiction and tautology.
---------------------------------------------
Now let's observe unbounded logical trees by using (without loss of generality) the 2-valued unbounded logical tree (no free variables are used):
This tree is also logically bounded by contradiction and tautology, but it is logically unbounded "below" (each one of its paths is an unbounded "below" distinct logical connective.
It is also observed that the diagonalisation argument can't be used along the 2-valued unbounded logical tree, since given any arbitrary unbounded logical path, its logical complement is already in this tree, which means that there are uncountable unbounded distinct logical paths along that tree.
-----------------------------
Now let's use the 2-valued unbounded logical tree in order to construct the natural numbers along it, by using the notion of radix point, as follows:
etc.
Please continue to read the second part of this post in http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/unbounded-logical-trees-2-a-19202.html#post87755
First, some background that leads to my question:
True is notated by 1
~True is notated by 0
p and q are two propositions as follows:
p = 0 0 1 1
q = 0 1 0 1
So, we get the 16 logical connectives as seen by the 16 distinct paths of the following binary tree:
Code:
p = 0 0 1 1
q = 0 1 0 1
--------------- /0 Contradiction
/0
/ \1 p AND q
/0
/ \ /0 p not implies q
/ \1
/ \1 p
/0
/ \ /0 q not implies p
/ \ /0
/ \ / \1 q
/ \1
/ \ /0 p XOR q
/ \1
/ \1 p OR q
*
\ /0 p NOR q
\ /0
\ / \1 p NXOR q
\ /0
\ / \ /0 NOT q
\ / \1
\ / \1 q implies p
\1
\ /0 NOT p
\ /0
\ / \1 p implies q
\1
\ /0 p NAND q
\1
\1 Tautology
The complements of a given binary tree with 16 distinct paths are:
Code:
p = 0 0 1 1
q = 0 1 0 1
--------------- /0 Contradiction -----------*
/0 |
/ \1 p AND q ---------------* |
/0 | |
/ \ /0 p not implies q -----* | |
/ \1 | | |
/ \1 p -----------------* | | |
/0 | | | |
/ \ /0 q not implies p -* | | | |
/ \ /0 | | | | |
/ \ / \1 q -------------* | | | | |
/ \1 | | | | | |
/ \ /0 p XOR q -----* | | | | | |
/ \1 | | | | | | |
/ \1 p OR q ----* | | | | | | |
* | | | | | | | |
\ /0 p NOR q ---* | | | | | | |
\ /0 | | | | | | |
\ / \1 p NXOR q ----* | | | | | |
\ /0 | | | | | |
\ / \ /0 NOT q ---------* | | | | |
\ / \1 | | | | |
\ / \1 q implies p -----* | | | |
\1 | | | |
\ /0 NOT p -------------* | | |
\ /0 | | |
\ / \1 p implies q ---------* | |
\1 | |
\ /0 p NAND q --------------* |
\1 |
\1 Tautology ---------------*
---------------------------------------------------
Let's briefly touch 3-valued logic.
True has 3 options which are: True, mTrue, ~True (m = middle, ~ = not).
True is notated by 2.
mTrue is notated by 1.
~True is notated by 0.
p, m and q are 3 propositions as follows:
Code:
p = 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
m = 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 2 2 2
q = 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2
A tree of 3-valued logic of these propositions has 327 = 7,625,597,484,987 logical connectives.
In this case contradiction is path 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0,
where tautology is path 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2.
Moreover, given any n-valued logical tree (where n > 1) it is bounded by contradiction and tautology.
---------------------------------------------
Now let's observe unbounded logical trees by using (without loss of generality) the 2-valued unbounded logical tree (no free variables are used):
Code:
*
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/ \
0 1
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \ / \
0 1 0 1
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
0 1 0 1 0 1 0 1
/ \ / \ / \ / \ / \ / \ / \ / \
/ \ / \ / \ / \ / \ / \ / \ / \
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
/ \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
. . .
This tree is also logically bounded by contradiction and tautology, but it is logically unbounded "below" (each one of its paths is an unbounded "below" distinct logical connective.
It is also observed that the diagonalisation argument can't be used along the 2-valued unbounded logical tree, since given any arbitrary unbounded logical path, its logical complement is already in this tree, which means that there are uncountable unbounded distinct logical paths along that tree.
-----------------------------
Now let's use the 2-valued unbounded logical tree in order to construct the natural numbers along it, by using the notion of radix point, as follows:
Code:
*
|\
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
0---------------1---------------Integers
|\ |\
| \ | \
| \ | \
| \ | \
| \ | \
| \ | \ Fractions
| \ | \
0 1 0 1
|\ |\ |\ |\
| \ | \ | \ | \
| \ | \ | \ | \
0 1 0 1 0 1 0 1
|\ |\ |\ |\ |\ |\ |\ |\
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
...
Code:
*
|\
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
| \
0 1
|\ |\
| \ | \
| \ | \
| \ | \
| \ | \
| \ | \
| \ | \
0-------1-------0-------1---------Integers
|\ |\ |\ |\
| \ | \ | \ | \
| \ | \ | \ | \ Fractions
0 1 0 1 0 1 0 1
|\ |\ |\ |\ |\ |\ |\ |\
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
...
etc.
Please continue to read the second part of this post in http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/unbounded-logical-trees-2-a-19202.html#post87755
Last edited: