• Is there any practical difference between logic / boolean algebra?
• Is Boolean algebra "algebra" because we have XYZ?

• COMP 2711H

  • 👨‍🏫Honors, so we will put infinite pressure on you HAHA
    • 0 coding, 100% theoretical
    • this session: easy (unlike others)
  • ~50%: have A's
    • but thou shalt suffer before getting it!
  • talking about basics: proof, proposition, set of natural numbers

• Propositional logic

  • proposition $:=$ any statement that can be true / false (not both)
    • 👨‍🏫or 1 or 0, as we are CS PPL
    • if you can't assign Boolean value: not a statement
      • e.g. "close the door" not boolean
    • usually using $p,q,r$ for var. names
    • atomic proposition prop. that can't be further broken down
    • for combination: use operators like $\neg$ ("neg")
      • and we often draw truth table to represent the value

• Logical operators

  • negation: $\neg p$ or $\bar{p}$

    • "it is not the case that $p$"

      • read as "not $p$"

    • (!p) ? 1 : 0

    • always opposite of $p$

      $p$	$\neg p$
      1	0
      0	1

    • examples

      • $p=$ "Prof. Amir uses Kali Linux for his desktop"
      • $\neg p=$ "Prof. Amir doesn't use Kali Linux for his desktop"

  • disjunction: $p\vee q$

    • "$p$ or $q$"

    • (p || q) ? 1 : 0

    • 0 only if both $p$ and $q$ are 0

      $p$	$q$	$p\vee q$
      1	1	1
      1	0	1
      0	1	1
      0	0	0

  • conjunction: $p\wedge q$

    • "$p$ and $q$"

    • (p && q) ? 1 : 0

    • 1 only if both $p$ and $q$ are T

    • 👨‍🏫 logical operator: kind of "promise" on condition

      $p$	$q$	$p\wedge q$
      1	1	1
      1	0	0
      0	1	0
      0	0	0

  • exclusive or (xor): $p\oplus q$

    • 1 when exactly one of $p$ and $q$ is T

      • 0 otherwise

    • ((p && !q) || (!p && q)) ? 1 : 0

      • or: (p ^ q) ? 1 : 0 (bit-wise though)

      $p$	$q$	$p\oplus q$
      1	1	0
      1	0	1
      0	1	1
      0	0	0

  • implication

    • cond. state.: $p\to q$

      • = prop. "if $p$ then $q$"

    • $p$: hypothesis / premise

    • $q$: conclusion / consequence / implication

    • (!p || (p && q)) ? 1 : 0

    • 👨‍🏫: any connection in $p,q$: not needed

      • simply depending on its values

    • 👨‍🏫: also, kind of promise

      • "if you are in classroom 2456, then you are Y1 student"

    • can be combined, like: $((p\wedge \neg q)\to r)\to t$

      $p$	$q$	$p\to q$
      1	1	1
      1	0	0
      0	1	1
      0	0	1

    • examples

      • if the moon is made of green cheese, then I have more money than Bill Gates
        • 👨‍🎓 as the hypothesis is false, the true value of conclusion doesn't matter
        • 👨‍🎓 you broke the promise (more like, premise) first!!!

  • 👨‍🏫 dual

    • let $s$ be a statement
    • $s^d$ is the statement obtained by following replacement $$\begin{gathered} \vee \to \wedge \\ \wedge \to \vee \\ T\to F \\ F\to T \\ \text{👨‍🎓 and }\neg \text{ reversed..?} \end{gathered}$$
    • example
      • $s=((x\vee y)\wedge \neg z)\wedge T$
      • $s^d=((x\wedge y)\vee \neg z)\vee F$
      • $s\equiv s^{\prime }⟺s^d\equiv s^{\prime d}$

• Conditional statement

  • necessary / sufficient conditions

    • if $(p\to q)==T$, then
      • $p$: sufficient condition of $q$
      • $q$: necessary condition of $p$
    • examples
      • if I am elected, then taxes will be lowered
      • if elected: sufficient to know "tax will be lowered"
      • if not "tax will be lowered" it must be that I am not elected
    • other ways for $p\to q$
      • if $p$, then $q$
      • $q$ if $p$
      • $q$ when $p$
      • $p$ implies $q$
      • $q$ follows from $p$
      • $p$ is a sufficient condition for $q$
      • $q$ is a necessary condition for $p$
      • $p$ only if $q$ (falsity of $q$: implies falsity of $p$)
        • $\neg q\to \neg p$

  • converse, contrapositive

    • (below: of $p\to q$)

    • converse: $q\to p$

    • contrapositive: $\neg q\to \neg p$

    • inverse: $\neg p\to \neg q$

    • $p\to q$: equivalent to its contrapositive

      $p$	$q$	$p\to q$	$\neg q\to \neg p$
      T	T	T	T
      T	F	F	F
      F	T	T	T
      F	F	T	T

  • bi-conditional statement

    • bicon. state.: $p\leftrightarrow q$

      • = prop. "$p$ if and only if $q$"
      • or: iff

    • T when $p$ and $q$ have the same truth value

      • or $(p\to q)\wedge (q\to p)$

    • equivalent

      • $p$ if and only if $q$
      • $p$ iff $q$
      • $p$ is "necessary and sufficient" for $q$
      • if $p$ then $q$, and conversely

      $p$	$q$	$p\leftrightarrow q$
      T	T	T
      T	F	F
      F	T	F
      F	F	T

• Tautology and contradiction

  • tautology: compound proposition, that is always true
    • no matter values of the prop. vars in it
      • 👨‍🏫: to proof: to show that certain statement is a tautology
  • contradiction: compound proposition, that is always false
  • contingency: compound proposition that is neither a tautology nor a contradiction
  • example
    • $p\vee \neg p$ is a tautology
      • as well as $(p\wedge \neg p)\to p$
      • as premise: always 0
    • $p\wedge \neg p$ is a contradiction
    • $p\to \neg p$ is a contingency

• Boolean algebra

  • conjunction, disjunction, or negation: can be implemented in hardware easily
    • 👨‍🏫 but, how can gates output true when all conjunctions are false?
      • for simplicity, just imagine there is an extra power line

graph TD;
      OR(( ∨ )) --> c_1[ ];
      a_1[ ] --> OR;
      b_1[ ] --> OR;
      AND(( ∧ )) --> c_2[ ];
      a_2[ ] --> AND;
      b_2[ ] --> AND;
      NOT(( ¬ )) --> c_3[ ];
      a_3[ ] --> NOT;

• let's build a simple adder!

  • bit 0 adder: $z_0=x_0\oplus y_0$
    • $\oplus$: defined as $(p\vee q)\wedge \neg (p\wedge q)$
    • read as, "XOR"
  • carry: $x_0\wedge x_0$
  • bit 1: $z_1=x_1\oplus y_1\oplus (x_0\wedge x_0)$
  • 👨‍🎓use 2611 knowledge
  • 👨‍🏫: one thing you can't ask: "where will this course be useful?"

• Logical equivalence

  • compound prop. always having the same truth values: logically equivalent
    • i.e. $p\leftrightarrow q$ is a tautology
    • notation: $p\equiv q$ (or $p\iff q$)
    • $\equiv$: not a logical operator
    • $p\equiv q$: not a compound prop., but a statement!
      • that $p\leftrightarrow q$ is a tautology
    • example: $p\to q\equiv \neg p\vee q$
      • 👨‍🎓 you can use truth table to compare! (manual)
  • use of $\equiv ,\leftrightarrow ,=$
    • $\equiv$: denotes logical equivalence, regardless of var. values
    • $\leftrightarrow$: binary logical operator
    • $=$: not used on propositions
      • only on two mathematical quantities

• Propositional equivalences

  • 👨‍🏫 many algebraic rules also work here!
    • consider $\vee$ as addition, and $\wedge$ as multiplication
  major equivalences

  Name	Equivalence
  Identity laws	$p\wedge T\equiv p$
  	$p\vee F\equiv p$
  Domination laws	$p\vee T\equiv T$
  	$p\wedge F\equiv F$
  Idempotent laws	$p\vee p\equiv p$
  	$p\wedge p\equiv p$
  Double negation laws	$\neg (\neg p)\equiv p$
  Commutative laws	$p\vee q\equiv p\wedge q$
  	$p\wedge q\equiv p\vee q$
  Associative laws	$(p\wedge q)\wedge r\equiv p\wedge (q\wedge r)$
  	$(p\vee q)\vee r\equiv p\vee (q\vee r)$
  Distributive laws	$p\vee (q\wedge r)\equiv (p\vee q)\wedge (q\vee r)$
  	$p\wedge (q\vee r)\equiv (p\wedge q)\vee (q\wedge r)$
  De Morgan's laws	$\neg (p\wedge q)\equiv \neg p\vee \neg q$
  	$\neg (p\vee q)\equiv \neg p\wedge \neg q$
  Absorption laws	$p\vee (p\wedge q)\equiv p$
  	$p\wedge (p\vee q)\equiv p$
  Negation laws	$p\vee \neg p\equiv T$
  	$p\wedge \neg p\equiv F$
  Involving cond. states	$p\to q\equiv \neg p\vee q$
  	$p\to q\equiv \neg q\to \neg p$
  	$(p\to q)\wedge (p\to r)\equiv p\to (q\wedge r)$
  	$(p\to q)\vee (p\to r)\equiv p\to (q\vee r)$
  	$(p\to r)\wedge (q\to r)\equiv (p\vee q)\to r$
  	$(p\to r)\vee (q\to r)\equiv (p\wedge q)\to r$
  Involving bicond. states.	$p\leftrightarrow q\equiv (p\to q)\wedge (q\to p)$
  	$p\leftrightarrow q\equiv \neg p\leftrightarrow \neg q$
  	$p\leftrightarrow q\equiv (p\wedge q)\vee (\neg p\wedge \neg q)$

• Examples

  • Use De Morgan's laws to express the negations of "Alice will send a secret message or Bob will send a secret message" and "today is Saturday and today is a holiday".
    • Alice & Bob
      • $p$ := Alice will send a secret message
      • $q$ := Bob will send a secret message
      • expr := $p\vee q$
      • $\neg$expr=$\neg (p\vee q)$=$\neg p\wedge \neg q$
      • = "Alice will NOT send a secret message and Bob will NOT send a secret message"
    • Today is...
      • $p$ := today is Saturday
      • $q$ := today is a holiday
      • expr := $p\wedge q$
      • $\neg$expr=$\neg (p\vee q)$=$\neg p\wedge \neg q$
  • Show that $\neg (p\to q)$ and $p\wedge \neg q$ are logically equivalent by developing a series of logical equivalences.
  1. $\neg (p\to q)$
  2. $p\wedge \neg q$
  • Show that $\neg (p\vee (\neg p\wedge q))$ and $\neg p\wedge \neg q$ are logically equivalent by developing a series of logical equivalences.
    1. $\neg (p\vee (\neg p\wedge q))$
    2. =$\neg p\wedge \neg q$
  • Show that $(p\vee q)\to (p\wedge q)$ is a tautology by developing a series of logical equivalences.
    1. $(p\vee q)\to (p\wedge q)$

• Rules of inference

  • theorem: statement that is always true (or: is a tautology)

  • we can write theorem as $h_1\wedge h_2\wedge \cdots \wedge h_n\to c$

  • better, we can define proof

    • you can either fill in $O(2^n)$ size table...
    • or: you use "inference" from existing tautology

  • proof for $h_1\wedge h_2\wedge \cdots \wedge h_n\to c$ is $\alpha$

    • a sequence of statements $$\begin{gathered} p_0:\dots \\ {p}_1:\dots \\ {p}_2:\dots \\ \dots \\ {p}_k=c \end{gathered}$$

  • s.t. for each $p_i$, either is true

    1. $p_i=h_j$ for some $j$
       • i.e. the premise
    2. $p_0\wedge p_1\wedge p2\wedge \cdots \wedge p_{i-1}\to p_i$ is a tautology
       • i.e. can be implied from prev. statements

  • 👨‍🏫 "one rule is actually enough"

    • but we introduce multiple for now

• Rules

  • modus ponens $$\begin{gathered} p \\ p\to q \\ --- \\ q \end{gathered}$$
    • thereafter, you can use $q$ as a tautology
    • or: "$(p\wedge (p\to q))\to q$ is a tautology"
    • 👨‍🏫 just another way of writing a tautology!
    • 👨‍🏫 the actual, only real rule here
  • syllogism $$\begin{gathered} p\to q \\ q\to r \\ --- \\ r \end{gathered}$$
    • or: "$((p\to q)\wedge (q\to r))\to r$" is a tautology
  • modus tollens $$\begin{gathered} p\to q \\ \neg q \\ --- \\ \neg p \end{gathered}$$