Proof of "Axioms" of Propositional Logic: Synopsis

[talanum1]

Here is the synopsis:

Proof of "Axioms" of Propositional Logic:
Synopsis.
Willem F. Esterhuyse.
Abstract.
We introduce more basic axioms with which we are able to prove some

"axioms" of Propositional Logic. We use the symbols from my other article:

"Introduction to Logical Structures". Logical Structures (SrL) are graphs with

doubly labelled vertices with edges carrying symbols. The proofs are very

mechanical and does not require ingenuity to construct. It is easy to see that in

order to transform information, it has to be chopped up. Just look at a kid playing

with blocks with letters on them: he has to break up the word into letters to

assemble another word. Within SrL we take as our "atoms" propositions with

chopped up relations attached to them. We call the results: (incomplete)

"structures". We play it safe by allowing only relations among propositions to be

choppable. We will see whether this is the correct way of chopping up sentences

(it seems to be). This is where our Attractors (Repulsors) and Stoppers come in.

Attractors that face away from each other repels and so break a relation between

the two propositions. Then a Stopper attaches to the chopped up relation to

indicate it can't reconnect. So it is possible to infer sentences from sentences. The

rules I stumbled upon, to implement this, seems to be consistent. Sources differ

asto the axioms they choose but some of the most famous "axioms" are proved.

Modus Ponens occurs in all systems.

1. Introduction.

We use new operators called "Attractors" and "Stoppers". An Attractor ( symbol:
"-(" OR ")-") is an edge with a half circle symbol, that can carry any relation
symbol. Axioms for Attractors include A:AA (Axiom: Attractor Annihilation)
where we have as premise two structures named B with Attractors carrying the
"therefore" symbol facing each other and attached to two neighboring structures:
B. Because the structures are the same and the Attractors face each other, and the
therefore symbols point in the same direction, they annihilate the structures B and
we are left with a conclusion of the empty structure. Like in:

((B)->-( )->-(B)) <-> (Empty Structure).

where "<->" means: "is equivalent to" or "follows from and vice versa".

A:AD reads as follows:

((A)->-(B))->-( <-> )->-(A) []->-(B)->-(

where "[]->-" is a Stopper carrying "therefore" relation.

We also have the axiom: A:AtI (Attractor Introduction) in which we have a row
of structures as premise and conclusion of the same row of structures each with an
Attractor attached to them and pointing to the right or left. Like in:

A B C D <-> (A)-( (B)-( (C)-( (D)-(

OR:

A B C D <-> )-(A) )-(B) )-(C) )-(D)

where the Attractors may carry a relation symbol.

Further axioms are: A:SD says that we may drop a Stopper at either end of a line.
And A:ASS says we can exchange Stoppers for Attractors (and vice versa) in a
line of structures as long as we replace every instance of the operators. A:AL says
we can link two attractors pointing trowards each other and attached to two
different structures, if the relations they carry is the same. A:SED says we may drop
an enclosure and Stopper carrying "AND" if this occurs at either end of a sentence or
any amount of encloseures (attached to Stoppers carrying "AND") if they all occur at
the end of a sentence.

A:AOA reads: (A)-(+)-( )-(+)-(A) <-> (A),

where "-(+)-" = "OR".

We prove Modus Ponens (T:MP) as follows:

Line nr. Statement Reason
1 B B -> C Premise
2 (B)->-( (B -> C)->-( 1, A:AtI
3 (B)->-( )->-(B) []->-(C)->-( 2, A:AD
4 []->-(C)->-( 3, A:AA
5 (C)->-( 4, A:SD
6 (C)->-[] 5, A:ASS
7 C 6, A:SD

We see that the Attractors cuts two structures into three (line 2 to line 3). In 2 "(B -> C)" is a structure.

We can prove AND-elimination, AND-introduction and transposition. We prove

Theorem: AND introduction (T:ANDI):

1 A B Premise
2 A -(x)-( B -(x)-( 1, A:AtI
3 (A)-(x)-[] (B)-(x)-[] 2, A:ASS
4 (A)-(x)-[] B 3, A:SD
5 (A)-(x)-( B 4, A:ASS
6 (A)-(x)-(B) 5, T:AL

where "-(x)-" = "AND", and T:AL is a theorem to be proved by reasoning
backwards through:

1 A -(x)- B Premise
2 A -(x)- B -(x)-( 1, A:AtI
3 &nbsp;)-(x)-(A) []-(x)-(B)-(x)-( 2, A:AD
4 []-(x)-(A) )-(x)-(B)-(x)-[] 3, A:ASS
5 A )-(x)-(B) 4, A:SD.

where the mirror image of this is proved similarly (by choosing to place the
Stopper on the other side of "-(x)-").

Modus Tollens and Syllogism can also be proven with these axioms.

We prove and-elimination: T:ANDE: (A)-(x)-(B) <> (A)

Proof:
1 (A)-(x)-(B) Premise
2 )-(x)-(A) []-(x)-(B)-(x)-( 1, A:AtI, A:AD
3 []-(x)-(A) )-(x)-(B)-(x)-[] 2, A:ASS
4 (A) )-(x)-(B) 3, A:SD
5 (A) []-(x)-(B) 4, A:ASS
6 (A) 5, A:SED

We prove: Theorem (T:O): (A OR A) -> A:

1 A -(+)- A Premise
2 (A)-(+)-((A)-(+)-(_)) 1, truth table
3 )-(+)-(A) []-(+)-((A)-(+)-(_))-(+)-( 2, A:AtI, A:AD
4 (A) []-(+)-((A)-(+)-(_)) 3, A:ASS, A:SD, A:ASS
5 (A)-(+)-( []-(+)- )-(+)-(A) []-(+)-(_)-(+)-( 4, A:AtI, A:AD
6 (A)-(+)-( )-(+)-(A) []-(+)-(_)-(+)-( 5, T:ANDE
7 (A) [](+)-(_)-(+)-( 6, A:AOA
8 A 7, A:EED

where "(_)" is the empty structure (a structure that is always false). Line 6 is because we can write line 5 as: (A)-(+)-( )-(+)-(A) []-(+)-(_)-(+)-( AND (A)-(+)-( []-(+)-(A) []-(+)-(_)-(+)-(, since they have the same meaning.

We prove Syllogism:

1 A -> B B -> C Premise
2 (A -> B)->-( (B -> C)->-( 1, A:AtI
3 &nbsp;)->-(A)->-[] (B)->-( )->-(B) []->-(C)->-( 2, A:ADx2
4 (A)->-[] (B)->-( )->-(B) []->-(C) 3, A:ASS, A:SDx2, A:ASS
5 (A)->-[] []->-(C) 4, A:AA
6 (A)->-( )->-(C) 5, A:ASS
7 A -> C 6, A:AL

For the Syllogism in the following form:

No B is C
All A is B

therefore

No A is C,

we need a variant on A:AA:

A:AA2: ((B)-->--( )-->--(No B))-->--(_) ... --(| X

where the dots mean the negation is introduced to the remaining sentence where the LS is removed, and the rightmost operator is an Introductor introducing negation.

Then we prove this as follows:

1 C -->-- No B B -->-- All A Premise
2 (C -->-- No B)-->--( (B -->-- All A)-->--( 1, A:AtI
3 (C)-->--[] (No B)-->--( )-->--(B) []-->--(All A)-->--( 2, A:AD
4 (C)-->--[] []-->--(All A)-->--( --(|X 3, A:AA2
5 (C)-->--( )-->--(All A)-->--( --(|X 4, A:ASS
6 (No C)-->--(All A)-->--( 5, A:AL
7 (No C)-->--(All A) 6, A:ASS, A:SD
8 (No A)-->--(C) 7, T:Transposition

For:

All B is Some C
All B is Some A

we see that A:AA would give us Some A is Some C as required, without having to draw a Venn diagram. Problem is: we can swop "All B" and "Some C" as required for this Syllogism by A:AA, but not if we model "is" as "therefore". If we model it by "therefore" we get "No C therefore Some A". This is where "is" doesn't allign with "therefore".

Of course the other two cases also holds as:

A:AA3: ((No B)-->--( )-->--(B))-->--(_) ...--(|X

A:AA4:((No B)-->--( )-->--(No B))-->--(_)

There remains the case:

All A is Some B
All B is All C

therefore

All A is Some C.

For this we need an axiom that includes "All C":

A:AA5: [(Some B)-->--( )-->--(All B) []-->--(All C)-->--( ]-->--(Some C).

where we used "[" instead of "(" since the other occurence would look like an Attractor.

For the syllogism of form:

All A is some B.
Some B is all C.

we see it does not follow that "All A is all C." since some B does not neccessarily mean the same as another instance of some B. The following follows though: "It is possible that all A is all C."

[talanum1]

Proof of T:Transposition

For a proof of T:Transposition see: "Academia.edu" and search for: "Proof of "Axioms" of Propositional Logic. WF Esterhuyse."

[talanum1]

I can prove another "axiom". As follows:

We need another axiom: A:SAL.

A:SAL: [ )-->--(A) ... []-->-- )-->--(A) ... ]--<->--[ )-->--(A) ... ... ]

where the intuition is: "Attractor linked to (A)" has been stopped, and the dots represent any strings of incomplete structures.

We also need:

A:AM: [ []-->--((A) []-->--(B))]--<->--[ []-->--(A) []-->-- []-->--(B)].

Now we can prove the axiom:

[(A)-->--((B)-->--(C))]--<->--[((A)-->--(B))-->--((A)-->--(C))]

by reasoning backwards through:

Proof:

Line # Statement Reason
1 ((A)-->--(B))-->--((A)->--(C)) Premise
2 )-->-- ((A)-->--(B))-->--((A)->--(C)) 1, A:AtI
3 )-->--((A)-->--(B)) []-->--((A)-->--(C))-->--( 2, A:AD
4 )-->-- )-->--((A)-->--(B) )-->-- []-->--((A)-->--(C))-->--( 3, A:AtI
5 )-->-- )-->--(A) []-->--(B)-->--( []-->-- )-->--(A) []-->-- []-->--(C)-->--( 4, A:AD, A:AM
6 )-->--(A) []-->--(B)-->--( []-->-- )-->--(A) []-->-- []-->--(C) 5, A:ASS, A:SD, A:ASS
7 )-->--(A) []-->--(B)-->--( []-->-- []-->--(C) 6, A:SAL
8 (A) []-->--(B)-->-- []-->-- []-->--(C) 7, T:AL
9 (A) []-->--((B)-->-- []-->-- []-->--(C)) 8, Choose Priority
10 (A) )-->--((B)-->-- )-->-- )-->--(C)) 9, A:ASS
11 (A)-->--((B)-->-- -->-- -->--(C)) 10, T:ALx3
12 (A)-->--((B)-->--(C)) 11, T:ANDEx2

[talanum1]

One can also prove "OR" introduction and "therefore" introduction from the premise A B. "A B" has the default relation "exist together" which has the same truth table as "AND" therefore these two follow.

[talanum1]

We should have stated: we may do a valid replacement of a substructure in a statement and state this on a new line, using the axioms.

[talanum1]

We have the axiom form Principia Mathematica:

(A)-->--(B) <> (((C)--(+)--(A))-->--((C)--(+)--(B))).

We could try to prove this by starting with the conclusion and reasoning backwards. However, we would need the axiom:

A:AAS: (C)--(+)--( []-->-- )--(+)--(C) <> []-->--

and this would contradict A:AOA.

Thus we need contradictive axioms to prove some axioms of Principia Mathematica. Does this mean some theorems of Principia does not follow? It seems to show there is a flaw in the basis of Principia.

[talanum1]

This axiom follows as is shown by a truth table, so we have to accept the axiom A:AAS and then ascribe some transformational idea to "[]-->--" so that A:AAS does not contradict A:AOA.

The flaw was therefore in my assumption of "[]-->--".

[talanum1]

(A)--(+)--(B) could also follow from the RS of the axiom.

[talanum1]

Can one use the objects: "class" and "set" interchangedly?

[talanum1]

Principia Mathematica has got a contradiction in it: definition *3.02 on p. 109 contradicts theorem: *3.34 on p. 112!

[talanum1]

Not so much a contradiction as a misalignment.

[talanum1]

Actually *3.02 is misaligned with *3.33!

[talanum1]

It seems that Principia Mathematica has another error: The deduction theorem states that if we have two lines in a proof: "A" and "B" we can conclude on another line: "A-->--B". However the inverse

hereof must not be valid since then we can prove: "(A)--(x)--(B)" from "(A)-->--(B)", which does not follow. This seems to be violated in Principia Mathematica: it derives:

"((Q)--<>--((P)--(x)--(S)))--<>--((Q)--<>--((P)--(x)--(R)))" from:

"((P)--<>--(Q))-->--((R)--<>--(S)) (1)" and:

"(((P)--(x)--(S))--<>--(P))--<>--(((P)--(x)--(S))--<>--(P)) (2)",

but now "(P)" is replaced with "(Q)" in LS and RS (2) and "(S)" with "(R)" in RS (2). This would seem to require us to be able to conclude "(P)--<>--(Q)" and "(R)--<>--(S)" on two separate lines since

either could be false in (1) while the whole sentence is true: this is contrary to the inverse of the deduction theorem not existing. Please reply if you agree (the proof may contain another replacement in

the sentence that formed (2)). The theorem was stated in terms of propositional functions, I just translated it into propositions.

[talanum1]

On p. 152 of Principia Mathematica it says (and I symbolize) ((x, y).W(x, y) into F(x, y))-->--((x, y).W(y, x) into F(x, y)) as a primitive proposition. Now the process "into" is not doable without exchanging x and y in F(x, y). PM assumes it is doable without exchanging x and y in F(x, y).

That means Mathematics assumes the impossible!

[talanum1]

The notation "(x, y).F(x, y)" that causes so much propositions to be proven involving distribution of "(x, y)" is obviated by my terminology in the paper: "Knowledge Organization" posted at Academia.edu. Search for "Willem F. Esterhuyse" together with this title.