# Section 1.2 - Tautology

The concept of a tautology is a function that always returns one. Essentially, each condition value of the function's domain will cause the function to return one. By my estimation, that makes the only function type that would cause conjunctive neglect to return one.

$$(\textbf{A}^n)_\alpha=1\ \forall\ (\textbf{A}^n)_\text{CV}\Rightarrow\big((\textbf{A}^n)_\alpha\big|_{(\textbf{A})_\text{CV}=0},\ldots,(\textbf{A}^n)_\alpha\big|_{(\textbf{A})_\text{CV}={2^n-1}}\big)_\cap\Leftrightarrow\big<(\textbf{A}^n)_\alpha\big>^{(\textbf{A}^n)}_\cap$$Naturally, if you're going to talk about when a function always goes right, you have to cover it all going wrong. This eternal wrongness is contradictory (now that's the basis for a religion). While not as obvious as the tautology/conjunctive neglect relationship, contradiction has a friend in disjunctive neglect. Contradiction says that every condition value of the function's domain will cause that function to return zero. Put differently, in a contradictory function there is no condition value, of the function's domain, that returns one.

$$(\textbf{A}^n)_\alpha=0\ \forall\ (\textbf{A}^n)_\text{CV}\Rightarrow \Big(\big((\textbf{A}^n)_\alpha\big|_{(\textbf{A}^n)_\text{CV}=0},\ldots,(\textbf{A}^n)_\alpha\big|_{(\textbf{A}^n)_\text{CV}=2^n-1}\big)_\cup\Big)_\mathtt{0x1}\Leftrightarrow \Big(\big<(\textbf{A}^n)_\alpha\big>^{(\textbf{A}^n)}_\cup\Big)_\mathtt{0x1}$$What I didn't expect were the opportunities for algebra. Exercise 1.5 through 1.29 were a grind of operator calculation. And, exercise 1.30 could be an example of creating compositions with explicit principal functions, term extraction, and switch statements. Once that came clear, the section was pretty cool.

#### Concepts and Vocabulary

**Tautology**: Always true. In practice, this is an application of conjunctive neglect.**Law of Excluded Middle**: $\big(A,(A)_\mathtt{0x1}\big)_\mathtt{0xE}$. The inclusion of every condition that makes $A$ true, and false.**Logically Imply**: If, then, for all condition values. $$\Big<\big((\textbf{A}^n)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0xD}\Big>^{(\textbf{A}^n)}_\cap$$**Logical Consequence**: Same as logically imply.**Logically Equivalent**: Two functions return the same state for all condition values. $$\Big<\big((\textbf{A}^n)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0x9}\Big>^{(\textbf{A}^n)}_\cap$$**Contradictory**: A truth function that returns zero for all condition values. $$\Big<\big((\textbf{A}^n)_\alpha\big)_\mathtt{0x1}\Big>^{(\textbf{A}^n)}_\cap$$**Logically True**: An actual sentence, in English (or something else), that can be uniquely translated into a tautology.**Logically False**: Like logically true, but a contraction.**Order of Operations**: $\neg$, $\land$, $\vee$, $\Rightarrow$, $\Leftrightarrow$.**Polish Notation**: A different way of writing connectives designed to eliminate parentheses and confuse people, while reinforcing a racial trope about eastern European methodology.**Duality**: This wasn't really explained, it was just the title of an exercise. The problem's text suggests that duality involves logical relationships between compositions of conjunctions and disjunctions, but the solutions, in the back, further imply that the problem was not accurately written.**Equivalent (Shannon, 1938)**: The same thing as logically equivalent, except with controls circuits, instead of truth functions.**Simpler (Shannon, 1938)**: A controls circuit with fewer contacts than an equivalent circuit.

- Proposition 1.1
- Proposition 1.2
- Proposition 1.3
- Proposition 1.4

$(\textbf{A}^n)_\alpha$ logically implies $(\textbf{A}^n)_\beta$ if $\Big<\big((\textbf{A}^n)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0xD}\Big>^{(\textbf{A}^n)}_\cap$

$(\textbf{A}^n)_\alpha$ is logically equivalent $(\textbf{A}^n)_\beta$ if $\Big<\big((\textbf{A}^n)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0x9}\Big>^{(\textbf{A}^n)}_\cap$.

If $(\textbf{A}^n)_\alpha=1\ \forall\ (\textbf{A}^n)_\text{CV}$ and $\Big<\big((\textbf{A}^n)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0xD}\Big>^{(\textbf{A}^n)}_\cap$ then $\big<(\textbf{A}^n)_\beta\big>^{(\textbf{A}^n)}_\cap$.

If $\big<(\textbf{A}^n)_\alpha\big>^{(\textbf{A}^n)}_\cap$ and $(\textbf{A}^n)_\beta=(\textbf{B}^n)_\alpha\ \ni\ B_i=(\textbf{A}^n)_{\gamma_i}$ then $\big<(\textbf{A}^n)_\beta\big>^{(\textbf{A}^n)}_\cap$.

$\bigg<\Big(\big((\textbf{A}^n)_\alpha,(\textbf{A}^n)_\beta\big)_\mathtt{0x9},\big((\textbf{B}^m)_\gamma,(\textbf{C}^m)_\gamma\big)_\mathtt{0x9}\Big)_\mathtt{0xD}\bigg>^{(\textbf{A}^n)}_\cap$ where some times when $(\textbf{A}^n)_\alpha$ is a member of $\textbf{B}^m$ it can be replaced by $(\textbf{A}^n)_\beta$