# Proof of "Axioms" of Propositional Logic.

## Recommended Posts

The file is attached.

##### Share on other sites

!

Moderator Note

Our rules state that members must be able to participate in a discussion without opening any docs or following any links. Is there any reason you can't copy/paste the information you want the members to have?

##### Share on other sites

I can't copy and paste here, because there are many figures in the text that does not show.

##### Share on other sites

2 hours ago, Willem F Esterhuyse said:

I can't copy and paste here, because there are many figures in the text that does not show.

Then give us your synopsis.

##### Share on other sites

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 are graphs with doubly labelled

vertices with edges carrying symbols. The proofs are very mechanical and does not

require ingenuity to construct.

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 where we have as premise two structures named B
with Attractors carrying the "therefore" symbol facing each other and attached to two
neighbouring 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).

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)-(.

A:AD distributes the Attractors and cut relations and places a Stopper on the cutted relation (see line 3 below). Stopper = "|-".
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 in a line of structures as long as
we replace every instance of the operators.

We can prove: P OR P = P. We prove Modus Ponens 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 can prove AND-elimination, AND-introduction and transposition. We prove
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    )-(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.

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

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

1    A -(+)- A                                    Premise
2    A -(+)- A  -(+)-(                        1, A:AtI
3    )-(+)-(A)  |-(+)-(A)-(+)-(          2, A:AD
4    |-(+)-(A)  )-(+)-(A)-(+)-|          3, A:ASS
5    A  )-(+)-(A)                                4, A:SD
6    A  |-(+)-(A)                                5, A:ASS

and from this (on introduction of a model taking only structures with truth tables as real)
we can conclude that A holds as required.

##### Share on other sites

• Phi for All unlocked this topic
• 2 weeks later...
Posted (edited)

Here is a better version:

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.

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".
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)

A:AD distributes the Attractors and cut relations and places a Stopper on the
chopped relation (see line 3 below). Stopper = "|-" or "-|".

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 towards each other and attached to two
different structures.

We can prove: P OR P -> P. We prove Modus Ponens 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).

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    )-(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.

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

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

1    A -(+)- A                                    Premise
2    A -(+)- A  -(+)-(                        1, A:AtI
3    )-(+)-(A)  |-(+)-(A)-(+)-(          2, A:AD
4    |-(+)-(A)  )-(+)-(A)-(+)-|          3, A:ASS
5    A  )-(+)-(A)                                4, A:SDx2
6    A  |-(+)-(A)                                5, A:ASS

and from this (on introduction of a model taking only structures with truth tables
as real) we can conclude that A holds as required.

We prove Syllogism:

1    A -> B  B -> C                                      Premise
2    (A -> B)->-(  (B -> C)->-(                    1, A:AtI
3    )->-(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

Edited by Willem F Esterhuyse
Spaces got included. Word got included.
##### Share on other sites

On 6/12/2022 at 8:36 PM, Willem F Esterhuyse said:

Here is a better version:

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.

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".
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)

A:AD distributes the Attractors and cut relations and places a Stopper on the
chopped relation (see line 3 below). Stopper = "|-" or "-|".

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 towards each other and attached to two
different structures.

We can prove: P OR P -> P. We prove Modus Ponens 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).

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    )-(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.

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

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

1    A -(+)- A                                    Premise
2    A -(+)- A  -(+)-(                        1, A:AtI
3    )-(+)-(A)  |-(+)-(A)-(+)-(          2, A:AD
4    |-(+)-(A)  )-(+)-(A)-(+)-|          3, A:ASS
5    A  )-(+)-(A)                                4, A:SDx2
6    A  |-(+)-(A)                                5, A:ASS

and from this (on introduction of a model taking only structures with truth tables
as real) we can conclude that A holds as required.

We prove Syllogism:

1    A -> B  B -> C                                      Premise
2    (A -> B)->-(  (B -> C)->-(                    1, A:AtI
3    )->-(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

Prove it...

##### Share on other sites

It has a justification for chopping up sentences added. Its a no-brainer.

##### Share on other sites

• 4 months later...

I run the logic in mind and as a result discovered places in mind where language starts. Now I think pages at a time.

##### Share on other sites

Here is the newest version of 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 neighbouring 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 vesa".

((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.

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

1    A -(+)- A                            Premise
2    A -(+)- A  -(+)-(                        1, A:AtI
3    )-(+)-(A)  []-(+)-(A)-(+)-(                    2, A:AD
4    []-(+)-(A)  )-(+)-(A)-(+)-[]                    3, A:ASS
5    A  )-(+)-(A)                            4, A:SDx2
6    A  []-(+)-(A)                            5, A:ASS

and from this (on introduction of a model taking only structures with truth tables
as real) we can conclude that A holds as required (structure A with a Stopper attached
to it does not have a truth table associated with it).

We prove Syllogism:

1    A -> B  B -> C                        Premise
2    (A -> B)->-(  (B -> C)->-(                    1, A:AtI
3    )->-(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

##### Share on other sites

7 minutes ago, Willem F Esterhuyse said:

Here is the newest version of 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 neighbouring 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 vesa".

((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.

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

1    A -(+)- A                            Premise
2    A -(+)- A  -(+)-(                        1, A:AtI
3    )-(+)-(A)  []-(+)-(A)-(+)-(                    2, A:AD
4    []-(+)-(A)  )-(+)-(A)-(+)-[]                    3, A:ASS
5    A  )-(+)-(A)                            4, A:SDx2
6    A  []-(+)-(A)                            5, A:ASS

and from this (on introduction of a model taking only structures with truth tables
as real) we can conclude that A holds as required (structure A with a Stopper attached
to it does not have a truth table associated with it).

We prove Syllogism:

1    A -> B  B -> C                        Premise
2    (A -> B)->-(  (B -> C)->-(                    1, A:AtI
3    )->-(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

##### Share on other sites

6 minutes ago, dimreepr said:

Like it. +1

##### Share on other sites

Also +1. And then I wanted to know from where you got that, and found one, that I am afraid, is even more realistic:

##### Share on other sites

An example is worth a thousand rules.

Can you give one?

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account