Propositional Calculus Proofs (5)

Th18) ⊢ P ⇒ (¬P ⇒ Q)
1.) ⊢ ¬P ⇒ (P ⇒ Q)                                            Th17
2.) ⊢  [¬P ⇒ (P ⇒ Q)] ⇒ [P ⇒ (¬P ⇒ Q)]           Th15
3.) ⊢ P ⇒ (¬P ⇒ Q)                                            MP 1,2

Th19) ⊢ ¬¬P ⇒ P
1.) ⊢ ¬¬P ⇒ (¬¬¬¬P ⇒ ¬¬P)                         Lk1
2.) ⊢ [¬(¬¬¬P) ⇒ ¬(¬P)] ⇒ (¬P ⇒ ¬¬¬P)   Lk3
3.) ⊢ [¬P ⇒ ¬(¬¬P)] ⇒ (¬¬P ⇒ P)                Lk3
4.) ⊢ ¬¬P ⇒ (¬¬P ⇒ P)                                   Th8
5.) ⊢ ¬¬P ⇒ P                                                  Th10

Th20) ⊢ Y ⇒ ¬¬Y

1.) ⊢ ¬¬(¬Y) ⇒ (¬Y)                                Th19

2.) ⊢ (¬(¬¬Y) ⇒ ¬Y) ⇒ (Y ⇒ ¬¬Y)         Lk3

3.) ⊢ Y ⇒ ¬¬Y                                           MP 1,2

Banner of Symfluence

Group One Symbols

Group Two Symbols

Group Three Symbols

Group Four Symbols

Unit Grid of Symfluence

Symbols of Order 0, 1, and 1/2

Faces/Symbols of Symfluence

Propositional Calculus Proofs (4)

Th17) ⊢ ¬T ⇒ (T ⇒ U)
1.) ⊢ (¬U ⇒ ¬T) ⇒ (T ⇒ U)                                  Lk3                               
2.) ⊢ ¬T ⇒ [(¬U ⇒ ¬T) ⇒ (T ⇒ U)]                   Th1
3.) ⊢ [¬T ⇒ (¬U ⇒ ¬T)] ⇒ [¬T ⇒ (T ⇒ U)]     Th2
4.) ⊢ ¬T ⇒ (¬U ⇒ ¬T)                                          Lk1
5.) ⊢ ¬T ⇒ (T ⇒ U)                                                MP 3,4

CIC Proofs (5)

Th12) ⊢ (P ⇒ Q) ⇒ [(Q ⇒ R) ⇒ (P ⇒ R)]
1.) (P ⇒ Q) ⇒ [(Q ⇒ R) ⇒ (P ⇒ Q)]                                    Lk1
2.) ⊢ (Q ⇒ R) ⇒ [(P ⇒ Q) ⇒ (P ⇒ R)]                                Th11
3.) ⊢ (P ⇒ Q) ⇒ {(Q ⇒ R) ⇒ [(P ⇒ Q) ⇒ (P ⇒ R)]}         Th1
4.) ⊢ (P ⇒ Q) ⇒ [(Q ⇒ R) ⇒ (P ⇒ R)]                                 Th9

Th13) A ⇒ (B ⇒ C), (D ⇒ B) ⊢ A ⇒ (D ⇒ C)

1.) ⊢ D ⇒ B                                                            Hyp
2.) ⊢ (D ⇒ B) ⇒ [(B ⇒ C) ⇒ (D ⇒ C)]              Th12

3.) ⊢ (B ⇒ C) ⇒ (D ⇒ C)                                     MP 1,2

4.) ⊢ A ⇒ (B ⇒ C)                                                Hyp

5.) ⊢ A ⇒ (D ⇒ C)                                                Th5

Th14) O ⇒ (U ⇒ V), (V ⇒ W) ⊢ O ⇒ (U ⇒ W)
 1.) ⊢ V ⇒ W                                                          Hyp
2.) ⊢ (V ⇒ W) ⇒ [(U ⇒ V) ⇒ (U ⇒ W)]           Th11
3.) ⊢ (U ⇒ V) ⇒ (U ⇒ W)                                   MP 1,2
4.) ⊢ O ⇒ (U ⇒ V)                                                Hyp
5.) ⊢ O ⇒ (U ⇒ W)                                               Th5

Th15) ⊢ [G ⇒ (H ⇒ J)] ⇒ [H ⇒ (G ⇒ J)]       Law of Commutation
1.) ⊢ [G ⇒ (H ⇒ J)] ⇒ [(G ⇒ H) ⇒ (G ⇒ J)]    Lk2
2.) ⊢ H ⇒ (G ⇒ H)                                              Lk1
3.) ⊢ [G ⇒ (H ⇒ J)] ⇒ [H ⇒ (G ⇒ J)]               Th13

Th16) ⊢ A ⇒ [(A ⇒ B) ⇒ B]                                Law of Assertion
1.) ⊢ (A ⇒ B) ⇒ (A ⇒ B)                                        Th3
2.) ⊢ [(A ⇒ B) ⇒ A] ⇒ [(A ⇒ B) ⇒ B]                 Th2
3.) ⊢ A ⇒ [(A ⇒ B) ⇒ A ]                                       Lk1
4.) ⊢ A ⇒  [(A ⇒ B) ⇒ B]                                       Th5

Cube

Formula Synthesis (Examples)

Assume D = [●,○,●,○] is the set of desired truth values for an unknown formula D. Since, D has four values, D must have two atomic components. Allow A and B to be atomic components of D. If A = [○,○,●,●] and B = [○,●,○,●], then the minterms of A and B are (A ∧ B) = [○,●,●,●] , (A ∧ ¬B) = [●,○,●,●] , (¬A ∧ B) = [●,●,○,●] , (¬A ∧ ¬B) = [●,●,●,○]. Consequently, (A ∧ ¬B) and (¬A ∧ ¬B) are the proper minterms to choose. When combined into a disjunction [(A ∧ ¬B) ∨ (¬A ∧ ¬B)] = [●,○,●,○] = D.

Assume C is an unknown formula and C = [○,●,●,●,●,●,●,●]. C must have three atomic components because there are eight truth values in C. If E, F, and G are the atomic components of formula C, then [(E ∧ F) ∧ G] = [○,●,●,●,●,●,●,●] is the only minterm needed because C has only one true truth value. C is what is called a "consensus" formula because E, F, and G must all be true in order for C to be true.

Back To Basics (5) Formula Synthesis

A logical formula possessing only one occurrence of ○ in its truth table is called a minterm.

The above examples can be used to construct logical formulae for any desired pattern of truth values given it has two components. Assume R is the desired yet unknown formula with two true and two false values and in the order. (Blogger is being stupid right now. I will have to put the truth values in row vector form... For some reason it won't let me upload another photo)

So, R = [○,○,●,●] and C1 = [○,●,●,●] C2 = [●,○,●,●] where C1 C2 are component formula. According to the above table C1 is the third column from the left and C2 is the fourth column from the left. Once the formulae are identified use a disjunction between the two component formula. C1 ∨ C2 or R ⇔ (A ∧ B) ∨ (¬A ∧ B). The formula is in what is called disjunctive normal form. It is a sequence of conjunctions divided by disjunctions.

Back to Basics (4) Proof by Contradiction

The method of proof by contradiction establishes a statement as true by showing that its negation leads to two mutually exclusive results. This method is made possible by the fact that in Boolean Algebraic logic a proposition is either true or false and no where in between. 

For every contradiction N and every proposition ᵹ the formula ᵹ ⇔ ( ¬ᵹ ⇒ N) is a tautology. If ¬ᵹ ⇒ N is true, then ᵹ is true by contraposition. N can also be represented by (ᴔ ∧ ¬ᴔ) where ᴔ is a proposition. If it can be shown that ¬ᵹ ⇒ ᴔ and ¬ᵹ ⇒ ¬ᴔ, then [¬ᵹ ⇒ (ᴔ ∧ ¬ᴔ)] ⇔ ( ¬ᵹ ⇒ N). Since, the proposition ¬ᵹ ⇒ (ᴔ ∧ ¬ᴔ) can be constructed (is true) the only possibility is that ¬ᵹ must be false.

Assume P and Q are propositions, P ⇒ Q, and (P ⇒ Q) ⇔ ᵹ. (P ⇒ Q) ⇔ ¬(P ∧ ¬Q). So, ¬ᵹ ⇔ (P ∧ ¬Q). In order to prove ¬ᵹ ⇒ (ᴔ ∧ ¬ᴔ), if P ⇒ ᴔ and ¬Q ⇒ ¬ᴔ, then ¬ᵹ ⇒ (ᴔ ∧ ¬ᴔ) since [(P ⇒ ᴔ) ∧ (¬Q ⇒ ¬ᴔ)] ⇒ [(P ∧ ¬Q) ⇒ (ᴔ ∧ ¬ᴔ)]. Therefore P ⇒ Q is true according to the method of proof by contradiction.