# CS代考 Propositional Logic – cscodehelp代写

Propositional Logic
COMP1600 / COMP6260
Australian National University
Semester 2, 2020

This Course
Programming. (Haskell, Java, Python, . . . )
Tell the computer how to perform a certain task
Logic. (this course)
Specify precisely what task should be performed
Computation. (this course)
the (discrete) maths of computation: what can be done in principle?
You know how to program. We will explore logic to describe programs, and study fundamental (programming language independent) models of what can be computed in the first place.
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Example: Vote Counting in the ACT
/* STEP 19 */
/* We haven’t incremented count yet, so this applies to NEXT count */ mark_elected(cand, (void *)(count+1));
/* STEP 20 */
vote_value_of_surplus = cand->c[count].total – quota;
/* STEP 21 */
increment_count();
/* STEP 22 */
pile = cand->c[cand->count_when_quota_reached].pile; non_exhausted_ballots
= (number_of_ballots(pile)
– for_each_ballot(pile, &is_exhausted, candidates));
/* STEP 23 */
if (non_exhausted_ballots == 0)
new_vote_value = fraction_one;
else
new_vote_value = ((struct fraction) { vote_value_of_surplus,
non_exhausted_ballots });
/* STEP 24 */
update_vote_values(pile, new_vote_value);
/* STEP 24b */
/* Report actual value (may be capped) */ report_transfer(count, pile->ballot->vote_value, vote_value_of_surplus); distribute_ballots(pile, candidates,vacating);
(Part of the EVACS code used for vote counting in the ACT in 2012)
Do you trust this? What does it do? Does it really implement the law? 3/35

The two faces of Boolean logic: Circuits
“Computation with boolean / binary values: 0 and 1”
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The two faces of Boolean logic: Formulae
George Boole (1815 – 1864)
“Logic” as a way of reasoning about true statements
at the time: mathematical statements
“All statements are either true orfalse: T /F ”
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Truth Values
Consider binary, or boolean values 0, or false (written F)
1, or true (written T)
Boolean Functions are functions that
take boolean values as inputs (zero, or more)
produce boolean values as outputs (generally, one or more)
Example.
x y f(x,y) x y f(x,y) 000 FFF 011 FTT 101 TFT 110 TTF
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Two Interpretations
Example.
x y f(x,y) x y f(x,y) 000 FFF 011 FTT 101 TFT 110 TTF
Computation with Binary Values
the output is 1 if precisely one of the inputs (but not both) is 1 or: considering x and y as one-bit numbers, the output is the sum x + y taken modulo two.
Truth of Statements
Here, x and y represent whether certain statements are true or false, e.g.: x could stand for “Joe was born in Melbourne”
y could stand for “Joe was born in Sydney”
Then f(x,y) is the truth value of the statement “Joe was born either in Sydney or in Melbourne”
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More on the Two Interpretations
Computing with Binary Values.
inputs are either 0 or 1
one output is prescribed for each combination used to describe computation
Truth of Statements
inputs describe whether certain statements are true or false output describes the truth value of a compound statement used to describe truth
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Building Boolean Functions
As with writing programs, boolean functions are constructed from basic building blocks
x AND y
x y x∧y 000 010 100 111
x OR y NOT x x y x∨y x ¬x 000 01 011 10 101
111
AND, OR and NOT are written as ∧, ∨ and ¬, respectively and called conjunction, disjunction and negation.
the symbols in the last row are circuit representations
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Example revisited
x y f(x,y) x y f(x,y) 000 FFF 011 FTT 101 TFT 110 TTF
Written as Logical Formulae
(x ∧¬y)∨(¬x ∧y)
Reading of the Formula. This formula is true in a situation
where either the LHS (i.e. x ∧ ¬y) is true, or the RHS (i.e. ¬x ∧ y) is true, or both (!)
theLHSistrueifx istrueand¬y istrue(soy isfalse) theRHSistrueif¬x istrue(sox isfalse)andy istrue
so that the formula is true in a situation where precisely one of x and y are true (but not both)
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Recall. Truth of the formula (x ∧ ¬y ) ∨ (¬x ∧ y ) depends on a situation that tells us which variables are true and false
can evaluate a formula to either true or false in any situation
Definition. If V is a set of variables (e.g. V = {x,y} for the formula above), then a situation for V is a mapping s : V → {T, F}
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Formula View
Q. How do we align the formula and the boolean function? A. Construct the value of the boolean function step by step
x y ¬x ¬y x∧¬y ¬x∧y (x∧¬y)∨(¬x∧y) FFTTFFF FTTFFTT TFFTTFT TTFFFFF
Left two columns: all possible truth values for x and y
given y, we know ¬y (4th column)
given x and ¬y , we know x ∧ ¬y (penultimate column)
given x, we know ¬x (3rd column)
given ¬x and y, we know ¬x ∧ y (6th column)
given x ∧¬y, and ¬x ∧y, we know (x ∧¬y)∨(¬x ∧y) (last column)
Indeed, (x ∧ ¬y ) ∨ (¬x ∧ y ) is exclusive-or!
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Formula View, Continued
(x ∧¬y)∨(¬x ∧y) Q. What columns do I need?
A. Look at the target formula . . .
to evaluate (x ∧¬y)∨(¬x ∧y), we need both x ∧¬yand ¬x ∧y to evaluate x ∧ ¬y , we need x (given) and ¬y
to evaluate ¬x ∧ y , we need ¬x and y (given)
Coloured formulae indicate columns in the table.
More systematically
1 start with the target formula, and create entries for immediate subformulae
2 repeat step 1 for all subformulae until inputs are reached.
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Truth Tables
Formulae. Are constructed from T (true), F (false) and variables (x, y, . . . ) using ∧, ∨, ¬ (more connectives later).
Truth Tables for a formula
list all T/F combinations of the variables (all situations) list the truth value of the formula, given variable values possibly list truth values of subformulae, too
Example. (same as before)
x y ¬x ¬y x∧¬y ¬x∧y (x∧¬y)∨(¬x∧y) FF FT TT TTFFFFF
F
T
T
F
F
T
T
F
F
T
F
F
T
T
F
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On Representation
Boolean Functions with n inputs are simply functions: f :{0,1}×···×{0,1}→{0,1}.
􏰣 􏰢􏰡 􏰤
n times
Logical Formula with n variables represent boolean functions the truth table of the formula defines a boolean function many representations of the same boolean function!
Circuits with n inputs also represent boolean functions
one boolean output is determined for each combination of inputs again, can have many circuits representing the same function
Summary. Boolean functions can be represented by truth tables (uniquely), formulae and circuits.
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More Circuit Primitives
x XOR y x NAND y
x y xor(x, y) x y nand(x, y) 000001 011011 101101 110110
(x ∧¬y )∨(¬x ∧y )
¬(x ∧ y)
we don’t introduce formula type connectives for NAND and XOR as they are not commonly used (and we will not use them in what follows), but they can be encoded
the symbols in the middle row are the circuit representations
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Can We Express All Boolean Functions?
Q. Given a boolean function f (x1, . . . , xn), can we construct a formula that represents f ?
A. Yes, this is a theorem and here is the proof. Recall that a formula represents a function if the truth table of the formula gives precisely the function table. construct formulae.
Proof (Sketch).
For boolean values i1,…,in ∈ {T,F} we have a formula φi1…in such that φi1…in(x1,…,xn) = T ⇐⇒ x1 = i1 and … and xn = in.
(This formula is a large ∧ with elements xl if il = T and ¬xl if il = F.)
Thentakeφf =φr1 ∨···∨φrk wherer1,…,rk arepreciselythevectors (i1 , . . . , in ) for which f (i1 , . . . , in ) = T .
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Proof Detail
Suppose we have a line of the truth table
x y z f(x,y,z)
Consider the formula then we have
TTF T φTTF ≡x∧y∧(¬z)
φTTF =T ifandonlyifx=T,y=T,z=F
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Proof Detail
Now consider all lines of the truth table
x y z f(x,y,z) …
TT

TT

FT

that have a T in the right hand column, where lines that are not shown have a F in the right column. The formula
φTTF ∨φTFF ∨φFFF has precisely the truth table above.
T
F
F
F
F
F
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Expressively Complete Sets
Definition. A set of logical connectives is expressively complete if it allows us to build all boolean functions.
Example. The set consisting of ∧, ∨ and ¬ expressively complete. Non-example. The sets consisting just of ¬, and just of ∨, are not
expressively complete.
Theorem. The set consisting of just nand is expressively complete.
Proof (Sketch). Using formula notation, we can use nand to express negation: ¬x = nand(x,x)
express conjunction: x ∧ y = ¬nand(x , y )
express disjunction: x ∨ y = ¬(¬x ∧ ¬y )
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Elements of Counting
Q1. How many boolean functions with n inputs (and one output) can we create using AND, OR and NOT circuits / using the logical connectives ∧, ∨, and ¬?
Q2. How many boolean functions with n inputs (and one output) can we create just with OR-gates and a constant wire with value 0 / the logical connectives ∨ and F?
Q3. How many gates do we need in the worst case to construct a boolean function?
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A Formula-Size Theorem
Theorem. Every boolean function with n input bits (and one output) can be represented using at most 10 · 2n − 10 logical connectives from the set {¬, ∧, ∨}.
Proof (Sketch). We use a technique called induction that we will analyse more closely later in the course.
Base case: n = 1. We can build every one-bit function using at most
10 · 21 − 10 = 10 connectives.
InductiveStep: n>1. Byfixingthefirstbitoff tobe0and1,wehave two n − 1-bit functions f0 and f1:
f0(x2,…,xn) = f(0,x2,…,xn) f1(x2,…,xn) = f(1,x2,…,xn)
Both can be represented using at most 10 · 2n−1 − 10 connectives.
We can write f = (¬x1 ∧ f0) ∨ (x1 ∧ f1). Using f0 and f1, we therefore need to add 4 more connectives to get a formula for f , and we have used
2·(10·2n−1 −10)+4=10·2n −20+4≤10·2n −10 connectives.
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Visualisation:
the boxes are the space of all situations where x and y are true or false labelled circles describe those situations where x and y are true
red area describes those situations where the formula is true.
(Every point in a Venn diagram describes a situation).
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Memories from High School . . .
Analogy: Algebraic Terms. Given a set V of variables, algebraic terms are constructed as follows:
0, 1, . . . and all variables x ∈ V are algebraic terms
if s and t are algebraic terms, then so is s +t and s ·t.
Example.
5+(3·x) x (x ·(3·(7+5)))+z Usual precedence. · binds more strongly than +:
x ·3+y reads as (x ·3)+y
Crucial Aspect.
Terms can be evaluated given values for all variables.
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Back to Boolean Functions
Definition. Given a set V of variables, boolean formulae are constructed as follows:
T (true) and F (false) and all variables x ∈ V are boolean formulae if φ and ψ are boolean formulae, then so are φ∧ψ and φ∨ψ.
if φ is a boolean formula, then so is ¬φ.
Examples.
T ∨¬(x ∨(¬y)) x x ∧¬(T ∨(F ∨x) Precedence. ¬ binds more strongly than ∧ binds more strongly than ∨:
Crucial Aspect.
¬x ∧y ∨z reads as ((¬x)∧y)∨z
Boolean formulae can be evaluated given (boolean) values for all variables.
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Equations
Examples from Algebra.
x · (3 + y) = x · 3 + x · y 25 + (18 · y) = 18 · y + 25
Boolean Equations. have boolean formulae on the left and right: x ∧ (y ∨ x) = x
T ∧ (y ∨ x) = x ∨ y
Valid Equations.
For all values of variables, LHS and RHS evaluate to same number. Applies to both algebraic terms and boolean formulae!
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Valid Boolean Equations.
Associativity
a ∨ (b ∨ c) = (a ∨ b) ∨ c a ∧ (b ∧ c) = (a ∧ b) ∧ c Commutativity
a∨b=b∨a a∧b=b∧a Absorption.
a ∨ (a ∧ b) = a a ∧ (a ∨ b) = a Identity.
a∨F=a a∧T=a Distributivity.
a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c)
Complements.
a ∨ ¬a = T a ∧ ¬a = F
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Equational Reasoning in Ordinary Algebra
(x +12)·3(x +17)
= (x +12)·(3·x +3·17)
= (x +12)·(3·x +51)
= x · (3 · x + 51) + 12 · (3 · x + 51)
= x · 3 · x + x · 51 + 12 · 3 · x + 12 · 51) = 3 · x2 + 51 · x + 36 · x + 612
= 3 · x2 + (51 + 36) · x + 612
= 3 · x2 + 87 · x + 612
(distributivity)
(distributivity) (distributivity) (commutativity) (distributivity)
each step (other than addition / multiplication of numbers) justified by a “law of arithmetic”
“pattern matching” against algebraic laws
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Proving Boolean Equations
Example. We prove the law of idempotence:
x ∨ x = (x ∨ x) ∧ T
= (x ∨ x) ∧ (x ∨ ¬x)
= x ∨ (x ∧ ¬x)
= x ∨ F x
Rules of Reasoning.
(identity) (complements) (distributivity) (complements) (identity)
All boolean equations may be assumed (with variables substituted by formulae)
may replace formulae with formulae that are proven equal equality is transitive!
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Two faces of boolean Equations
Truth of boolean equations:
A boolean equation φ = ψ (where φ, ψ are boolean formulae) is true if φ and ψ evaluate to the same truth values in all situations (i.e. for all possible truth values of the variables that occur in φ and ψ).
Equational Provability of boolean equations:
A boolean equation is provable if it can be derived from associativity, commutativity, absorption, identity, distributivity and complements using the laws of equational reasoning.
Q. How do these two notions hang together?
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Soundness and Completeness
Slightly Philosophical.
Truth of an equation relates to the meaning (think: truth tables) of the connectives ∧, ∨ and ¬.
Equational provability relates to a method that allows us to establish truth of an equation.
They are orthogonal and independent ways to think about equations. Soundness. If a boolean equation φ = ψ is provable using equations, then
it is true.
all basic equations (associativity, distributivity, . . . ) are true the rules of equational reasoning preserve truth.
Completeness. If a boolean equation is true, then it is provable using equations.
more complex proof (not given here), using the so-called Lindenbaum Construction.
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Challenge Problem: The De
De Morgan’s Laws
¬(x ∨ y) = ¬x ∧ ¬y ¬(x ∧ y) = ¬x ∨ ¬y
In English
if it is false that either x or y is true, they must both be false
if it is false that both x and y are true, then one of them must be
false.
Truth of De Morgan’s Laws: Easy to establish via truth tables.
Provability of De Morgan’s Laws
if the completeness theorem (that we didn’t prove!) is true, then an
equational proof must exist
however, it is quite difficult to actually find it!
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