Set Theory and Symfluence (0)

The symbol ' ∈ ' has a measurable effect on the mind. 

Exercise 1
Look at the symbol ' ∈ ' and in your mind's voice say ' is an element of '. Repeat!

New possibilities now exist within your mind. For example:  A ∈ B.

Exercise 2
Look at the symbol ' ¬ ' and in your mind's voice say 'It is not the case that'. Repeat!

Abstractions such as '¬ A ∈ B' and ' X ∈ Y ' select very specific thoughts for your conscious experience. They're symfluence or symbol-influence finds its origin in the English language and is strengthened with each repetition of exercises 1 and 2.

Exercise 3

Look at the symbol ' ∧ ' and in your mind's voice say 'and'. Repeat!

It is not the case that the English language has provided you with all of the patterns of thought necessary to understand set theory. Thoughts such as ' A ∈ B ∧ B ∈ C ' are perfectly well understood now without any further training. However, making all of the abstractions of set theory available to your conscious experience will require a new training method.

Examples of Proofs in Philosophical Logic

B, (C ∧ ¬D) ⊢ B ∧ ¬D
1 B                          Assumption
2 C ∧ ¬D                Assumption
3 ¬D                       2 ∧E
4 B ∧ ¬D                1,3 ∧I

[B ⇔ (D ⇔ ¬A)], (B ∧ D) ⊢ [¬A ∧ (B ∨ C)]
1 B ⇔ (D ⇔ ¬A)         Assumption
2 B ∧ D                        Assumption
3 B                                2 ∧E
4 D ⇔ ¬A                    1,3  ⇔E
5 D                               2 ∧E
6 ¬A                             4,5 ⇔E
7 B ∨ C                         3 ∨I
8 ¬A ∧ (B ∨ C)             6,7 ∧I

Philosophical Propositional Calculus (1)

Subderivaton Rules

Each subderivation begins with an auxiliary assumption.

Conditional Introduction (⇒I) If P is an assumption that leads to Q, then P implies Q.

Negation Introduction (¬I) If an assumption P leads to both Q and not Q, then not P must be true.

Negation Elimination (¬E) If an auxiliary assumption P leads to Q and not Q, then P.

Disjunction Elimination (∨E) If P ∨ Q, and P leads to R and Q leads to R, then R.

Biconditional Introduction (⇔I) If P leads to Q and Q leads to P, then Q ⇔ P.

Philosophical Propositional Calculus (0)

Derivation Rules:

Reiteration (R) In a proof, a previously established result can always be stated again.
P
P

Conjunction Elimination (∧E) When two statements are conjoined each component can be stated seperately.
P ∧ Q            P ∧ Q
P                   Q

Conjunction Introduction (∧I) Two seperate results can always be conjoined as a conjunction.
P
Q
P ∧ Q

Disjunction Introduction (∨I) For any result P, P or some other statement is always true.
P                      P
P ∨ Q               Q ∨ P

Conditional Elimination (⇒E) The rule of Modus Ponens (MP)
P ⇒ Q
P
Q

Biconditional Elimination (⇔E) If two statements are equivalent and either is true, then the other must too be true.
P ⇔ Q               P ⇔ Q
P                        Q
Q                        P

A Change In Plans

I discovered some errors in the textbook that I've been working through. I now distrust the predicate logic presented there. All of the mathematical theories are correct so far as I can tell. So, I shall return to it after I establish a strong understanding of predicate calculus through other textbooks. Philosophical logic is an excellent method to learn prior to learning mathematical logic. So, I shall approach predicate calculus from this angle first as I did with propositional logic.

Set Theory (0)

The axiom of the empty set ∃y{∀x[¬(x ∈ y)]} asserts there exists a set that contains no elements. Ever seen an empty bag? The empty set in mathematics is represented by either { } or Ø.

The axiom of extensionality ∀A{∀B[∀x(x ∈ A ⇔ x ∈ B) ⇔ ∀y(A ∈ y ⇔ B ∈ y)]} claims that if you find two sets each with the same elements, then you will find each set contained by precisely the same sets. Basically, ∀x(x ∈ A ⇔ x ∈ B) and ∀y(A ∈ y ⇔ B ∈ y) are long winded versions of A = B. With the axiom of extensionality it can be proved that every set equals itself. In other words, ⊨ ∀S(S = S).

ST1.) ⊨ ∀S(S = S)

1.) ⊢ P ⇔ P                                       Th54

2.) ⊨ x ∈ S ⇔ x ∈ S                          Substitution

3.) ⊨ ∀S[∀x( x ∈ S ⇔ x ∈ S )]        Gen

4.) ⊨ ∀S(S = S)                                 Definition of =

Definition of ⊆

⊨ ∀A{∀B[(A ⊆ B) ⇔ {∀x[(x ∈ A) ⇒ (x ∈ B)]}]}

ST2.) ⊨ ∀S(Ø ⊆ S)

1.) ⊢ F ⇒ P                                       tautology with F false

2.) ⊨ x ∈ Ø ⇒ x ∈ S                         Substitution

3.) ⊨ ∀S[∀x( x ∈ Ø ⇒ x ∈ S )]        Gen

4.) ⊨ ∀S(Ø ⊆ S)                              Definition of ⊆

Predicate Calculus Proofs (1)

Pr1.) x does not occur free in P, ∀x(P ⇒ Q) ⊨ P ⇒ ∀x(Q)
1.)  ⊨ ∀x(P ⇒ Q)                                                                       Hyp
2.)  ⊨ [∀x (P ⇒ Q)] ⇒ {P ⇒ [∀x(Q)]}                                      Z5
3.)  ⊨ P ⇒ ∀x(Q)                                                                       MP 1,2

Pr2.) ⊨ ∃x(P) ⇔ [¬∀x(¬P)]
1.) ⊨ ¬∃x(P) ⇔ ∀x(¬P)                                              Z6
2.) ⊨ ¬[¬∃x(P)] ⇔ ¬∀x(¬P)                                     Th53
3.) ⊨ ¬∀x(¬P) ⇔ ¬[¬∃x(P)]                                     Th41
4.) ⊨ ¬[¬∃x(P)] ⇔ ∃x(P)                                          Th49
5.) ⊨ ¬∀x(¬P) ⇔ ∃x(P)                                             Th42
6.) ⊨  ∃x(P) ⇔ ¬∀x(¬P)                                            Th41

Pr3.) ⊨ ∀x(P) ⇔ [¬∃x(¬P)]
1.) ⊨ [¬∀x(P)] ⇔ ∃X(¬P)                        Z7
2.) ⊨ {¬[¬∀x(P)]} ⇔ [¬∃X(¬P)]            Th53
3.) ⊨  [¬∃X(¬P)] ⇔ {¬[¬∀x(P)]}            Th41
4.) ⊨ {¬[¬∀x(P)]} ⇔ ∀x(P)                     Th49
5.) ⊨ [¬∃X(¬P)] ⇔ ∀x(P)                        Th42
6.) ⊨ ∀x(P) ⇔ [¬∃X(¬P)]                        Th41

Pr4.) ⊨  ∀x(P ⇒ Q) ⇒ [∀x(P) ⇒ Q]

1.) ⊨ ∀x(P ⇒ Q) ⇒ (P ⇒ Q)                      Z4

2.) ⊨ ∀x(P) ⇒ P                                         Z4
3.) ⊨ ∀x(P ⇒ Q) ⇒ [∀x(P) ⇒ Q]              Th13