Learning Goals: Pre-Class Learning Goals: In-Classcs-121/2013W2/... · • Prelude: What Is Proof?...

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CPSC 121: Models of Computation

2013W2

Proof (First Visit)

Steve Wolfman, based on notes by Patrice Belleville, Meghan Allen and others

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This work is licensed under a Creative Commons Attribution 3.0 Unported License.

Outline

• Prereqs, Learning Goals, and Quiz Notes

• Prelude: What Is Proof?

• Problems and Discussion

– “Prove Your Own Adventure”

– Why rules of inference? (advantages + tradeoffs)

– Onnagata, Explore and Critique

• Next Lecture Notes

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Learning Goals: Pre-Class

By the start of class, you should be able to:

– Use truth tables to establish or refute the

validity of a rule of inference.

– Given a rule of inference and propositional

logic statements that correspond to the rule’s

premises, apply the rule to infer a new

statement implied by the original statements.

3

Learning Goals: In-Class

By the end of this unit, you should be able to: – Explore the consequences of a set of

propositional logic statements by application of equivalence and inference rules, especially in order to massage statements into a desired form.

– Critique a propositional logic proof; that is, determine whether or not is valid (and explain why) and judge the applicability of its result to a specific context.

– Devise and attempt multiple different, appropriate strategies for proving a propositional logic statement follows from a list of premises.

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Where We Are in

The Big Stories

Theory

How do we model

computational systems?

Now: Continuing to build the

foundation for our proofs.

(We’ll get to the level of

proof we really need

starting with the next

unit.)

Hardware

How do we build devices to

compute?

Now: Taking a bit of a

vacation in lecture!

5

Motivating Problem: Changing cond Branches

Assuming that a and c cannot both be true and that this function produces true:

;; Boolean Boolean Boolean Boolean -> Boolean

(define (rearrange-cond? a b c d)

(cond [a b]

[c d]

[else e]))

Prove that the following function also produces true:



(cond [c d]

[a b]

[else e]))

6

But first, prove these handy “lemmas”:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

(Reality check: you must be

able to do formal proofs. But,

as with using equivalence laws

to reorganize code, in practice

you’ll often reason using proof

techniques but without a formal

proof.)

http://creativecommons.org/licenses/by/3.0/http://creativecommons.org/licenses/by/3.0/http://creativecommons.org/licenses/by/3.0/

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NOT a Quiz Note

~p

~(p v q)

a. This is valid by generalization (p p v q).

b. This is valid because anytime ~p is true, ~(p v q) is also true.

c. This is invalid by generalization (p p v q).

d. This is invalid because when p = F and q = T, ~p is true but ~(p v q) is false.

e. None of these.

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What does this mean?

We can always substitute something equivalent for a subexpression of a logical expression.

We cannot always apply a rule of inference to just a part of a logical statement.

Therefore, we will only apply rules of inference to complete statements, no matter what!

11

Outline








12

What is Proof?

A rigorous formal argument that

unequivocally demonstrates the

truth of a proposition, given the truth

of the proof’s premises.

Adapted from MathWorld: http://mathworld.wolfram.com/Proof.html 13

What is Proof?

A rigorous formal argument that

unequivocally demonstrates the

truth of a proposition (conclusion),

given the truth of the proof’s

premises.

Adapted from MathWorld: http://mathworld.wolfram.com/Proof.html 14

Problem: Meaning of Proof

Let’s say you prove the following:

Premise 1

Premise 2

⁞

Premise n

Conclusion

Can one of the premises be false?

a. No, proofs may not use false premises

b. No, the proof shows that the premises are true

c. Yes, but then the conclusion is false

d. Yes, but then we know nothing about the conclusion

e. Yes, but we still know the conclusion is true 15

http://mathworld.wolfram.com/Proof.htmlhttp://mathworld.wolfram.com/Proof.html

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Tasting Powerful Proof:

Some Things We Might Prove • We can build a “three-way switch” system with any

number of switches.

• We can build a combinational circuit matching any truth table.

• We can build any digital logic circuit using nothing but NAND gates.

• We can sort a list by breaking it in half, and then sorting and merging the halves.

• We can find the GCD of two numbers by finding the GCD of the 2nd and the remainder when dividing the 1st by the 2nd.

• Is there any fair way to run elections?

• Are there problems that no program can solve?

Meanwhile... 16

What Is a Propositional Logic

Proof?

An argument in which:

(1) each line is a propositional logic statement,

(2) each statement is a premise or follows

unequivocally by a previously established

rule of inference from the truth of previous

statements, and

(3) the last statement is the conclusion.

A very constrained form of proof, but a good starting point.

Interesting proofs will usually come in less structured

packages than propositional logic proofs. 17

Outline








18

Prop Logic Proof Problem

To prove:

~(q r)

(u q) s

~s ~p___

~p

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“Prove Your Own Adventure”

To prove:

~(q r)

(u q) s

~s ~p___

~p

Which step is the easiest to fill in?

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise

[STEP A: near the start]

[STEP B: in the middle]

[STEP C: near the end]

[STEP D: last step]

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D: Last Step

To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise

...

~q ~r De Morgan’s (1)

~q Specialization (?)

...

((u q) s) Bicond (2)

(s (u q))

... ~s

~p Modus ponens (3,?)

Why do we want to put ~p at the end?

a. ~p is the proof’s conclusion

b. ~p is the end of the last premise

c. every proof ends with ~p

d. None of these but some other reason

e. None of these because we don’t

want it there

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4

C: Near the End

To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise

...



...


(s (u q))

... ~s


Why do we want to put the blue

line/justification at the end?

a. ~s ~p is the last premise

b. ~s ~p is the only premise that

mentions ~s

c. ~s ~p is the only premise that

mentions p

d. None of these but some other reason

e. None of these b/c we don’t want it there 22

A: Near the Start

To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise

...



...


(s (u q))

... ~s


Why do we want the blue

lines/justifications?

a. ~(q r) is the first premise

b. ~(q r) is a useless premise

c. We can’t work directly with a premise

with a negation “on the outside”

d. Neither the conclusion nor another

premise mentions r

e. None of these 23

B: In the Middle

To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise

...



...


(s (u q))

... ~s


Why do we want the blue

line/justification?

a. (u q) s is the only premise left

b. (u q) s is the only premise that

mentions u

c. (u q) s is the only premise that

mentions s without a negation

d. We have no rule to get directly from

one side of a biconditional to the other

e. None of these

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Prop Logic Proof Strategies

• Work backwards from the end

• Play with alternate forms of premises

• Identify and eliminate irrelevant information

• Identify and focus on critical information

• Alter statements’ forms so they’re easier to work with

• “Step back” from the problem frequently to think about assumptions you might have wrong or other approaches you could take

And, if you don’t know that what you’re trying to prove follows... switch from proving to disproving and back now and then.

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Continuing From There

To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise

4. ~q ~r De Morgan’s (1)

5. ~q Specialization (4)

6. ((u q) s) Bicond (2)

(s (u q))

7. ?????? Specialization (6) ...

~s


Which direction of goes in step 7?

a. (u q) s because the simple part

is on the right

b. (u q) s because the other

direction can’t establish ~s

c. s (u q) because the simple part

is on the left

d. s (u q) because the other

direction can’t establish ~s

e. None of these

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Aside: What does it mean to

“work backward”? Take the conclusion of the proof.

Use a rule in reverse to generate something

closer to a statement you already have (like

a premise).

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5

Finishing Up (1 of 3)

To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise



6. ((u q) s) Bicond (2)

(s (u q))

7. s (u q) Specialization (6)

8. ???? ???? 9. ~(u q) ????

10. ~s Modus tollens (7, 9)

11. ~p Modus ponens (3,10)

We know we needed ~(u q) on

line 9 because that’s what we

created line 7 for!

Side Note: Can we work directly

with a statement with a negation

“on the outside”?

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To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise



6. ((u q) s) Bicond (2)

(s (u q))


8. ???? ???? 9. ~(u q) ????



We know we needed ~(u q) on

line 9 because that’s what we

created line 7 for!

Now, how do we get ~(u q)?

Working forward is tricky. Let’s

work backward. What is ~(u q)

equivalent to? 39


To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise



6. ((u q) s) Bicond (2)

(s (u q))


8. ~u ~q ???? 9. ~(u q) De Morgan’s (8)



All that’s left is to get to ~u ~q.

How do we do it?

40


To prove:

~(q r)

(u q) s

~s ~p___

~p

1. ~(q r) Premise

2. (u q) s Premise

3. ~s ~p Premise



6. ((u q) s) Bicond (2)

(s (u q))


8. ~u ~q Generalization (5) 9. ~(u q) De Morgan’s (8)



As usual in our slides, we made no

mistakes and reached no dead

ends. That’s not the way things

really go on difficult proofs!

Mistakes and dead ends are part of

the discovery process! So, step

back now and then and reconsider

your assumptions and approach! 41

Outline








42

Limitations of Truth Tables

Why not just use truth tables to prove propositional logic theorems?

a. No reason; truth tables are enough.

b. Truth tables scale poorly to large problems.

c. Rules of inference and equivalence rules can prove theorems that cannot be proven with truth tables.

d. Truth tables require insight to use, while rules of inference can be applied mechanically.

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Limitations of

Logical Equivalences Why not use logical equivalences to prove that

the conclusions follow from the premises?

a. No reason; logical equivalences are enough.

b. Logical equivalences scale poorly to large problems.

c. Rules of inference and truth tables can prove theorems that cannot be proven with logical equivalences.

d. Logical equivalences require insight to use, while rules of inference can be applied mechanically.

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Outline






– Onnagata: Explore and Critique


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Preparatory Comments

When we apply logic to a domain, we give interpretations

for the logical symbols. That interpretation is where we can

argue things like “meaning”, “values”, and “moral right”.

Within the logical context, we argue purely on the basis of

structure and irrefutable manipulations of that structure.

And… statements contradict each other when, taken

together, they are logically equivalent to F, such as (a ~a). There is no way for them to be simultaneously

true.

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Problem: Onnagata

Problem: Critique the following argument.

Premise 1: If women are too close to femininity to portray

women then men must be too close to masculinity to

play men, and vice versa.

Premise 2: And yet, if the onnagata are correct, women are

too close to femininity to portray women and yet men

are not too close to masculinity to play men.

Conclusion: Therefore, the

onnagata are incorrect, and

women are not too close to

femininity to portray women.

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Contradictory Premises?

Do premises #1 and #2 contradict each other (i.e., is

(premise1 premise2) logically equivalent to F)?

a. Yes

b. No

c. Not enough information to tell.

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Defining the Problem

Does it make sense to use the definition “w = women” for a

propositional logic variable w?

a. Yes, in this problem.

b. Yes, but not in this problem.

c. No, not in this problem.

d. No, not in any problem.

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Translating the Statements

Which of these is an accurate translation of

one of the statements?

a. w m

b. (w m) (m w)

c. o (w ~m)

d. ~o ~w

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Contradictory Premises?

So premises #1 and #2 are w m and o (w ~m).

Do premises #1 and #2 contradict each other (i.e., is

(premise1 premise2) logically equivalent to F)?

a. Yes

b. No

c. Not enough information to tell.

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Problem: Now, Explore!

Critique the argument by either:

(1) Proving it correct (and commenting on how good the propositional logic model’s fit to the context is). How do we prove prop logic statements?

(2) Showing that it is an invalid argument. How do we show an argument is invalid? (Remember the quiz!)

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Outline








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Next Lecture Learning Goals:

Pre-Class By the start of class, you should be able to:

– Evaluate the truth of statements that include predicates applied to particular values.

– Show predicate logic statements are true by enumerating examples (i.e., all examples in the domain for a universal or one for an existential).

– Show predicate logic statements are false by enumerating counterexamples (i.e., one counterexample for universals or all in the domain for existentials).

– Translate between statements in formal predicate logic notation and equivalent statements in closely matching informal language (i.e., informal statements with clear and explicitly stated quantifiers).

55

Next Lecture Prerequisites

Review (Epp 4th ed) Chapter 2 and be able

to solve any Chapter 2 exercise.

Read Sections 3.1 and 3.3 (skipping the

“Negation” sections in 3.3)

Complete the open-book, untimed online

quiz.

56

8


Assuming that a and c cannot both be true and that this function produces true:



(cond [a b]

[c d]

[else e]))

Prove that the following function also produces true:



(cond [c d]

[a b]

[else e]))

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First, prove these handy “lemmas”:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)


Assuming that a and c cannot both be true, and that this function produces true:



(cond [a b]

[c d]

[else e]))

We leave the lemmas as an exercise:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

In prop logic:

1. ~(a b) premise

2. (a b) (~a ((c d) (~c e))) premise

3. …

4. (c d) (~c ((a b) (~a e))) target conclusion

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We’ll use our “heuristics” to

work forward and backward

until we solve the problem.


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. …

4. (c d) “subgoal”

5. (~c ((a b) (~a e))) “subgoal”

6. (c d) (~c ((a b) (~a e))) by CONJ on 4, 5

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Lemmas:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

We start by working backward;

how de we prove x y? Well,

one way is to prove x and also

prove y. We’ll break those into

two separate subproblems!

Side note: we’ll use the two statements

you proved as exercises as “lemmas”:

rules we proved for use in this proof.

(Want to use them on an assignment /

exam? Prove them there!)


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. …


5. (~c (a b)) (~c (~a e))) “subgoal”

6. (~c ((a b) (~a e))) Lemma 1 on 5


60

Lemmas:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

The second of these subgoals is

still huge. We decided to break

it into two pieces (and that’s

why we went off and proved

Lemma 1).


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. …


5. ~c (a b) “subgoal”

6. ~c (~a e) “subgoal”

7. (~c (a b)) (~c (~a e))) by CONJ on 5, 6

8. (~c ((a b) (~a e))) Lemma 1 on 7


61

Lemmas:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

Now, we can attack those two

pieces separately (which feels like it

might be the wrong approach to

me… but worth a try!)


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. …




8. (~c (a b)) (~c (~a e))) by CONJ on 6, 7

9. (~c ((a b) (~a e))) Lemma 1 on 8


62

Lemmas:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

I’m out of ideas at the end. I switch to the beginning

and play around with premises. (Foreshadowing: I

didn’t figure out what to do with this premise until

near the end.)

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In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. …




9. (~c (a b)) (~c (~a e))) by CONJ on 7, 8

10. (~c ((a b) (~a e))) Lemma 1 on 9


63

Lemmas:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

Let’s try the other premise.


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2

6. …




10. (~c (a b)) (~c (~a e))) by CONJ on 8, 9

11. (~c ((a b) (~a e))) Lemma 1 on 10


64

Lemmas:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

Continuing with that premise…

Hey! We can use our Lemma again!


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2

6. (~a (c d)) (~a (~c e)) by Lemma 1 on 5

7. …




11. (~c (a b)) (~c (~a e))) by CONJ on 9, 10

12. (~c ((a b) (~a e))) Lemma 1 on 11


65

Lemmas:

1. p (q r) (p q) (p r)

2. p (q r) q (p r)

Continuing with

that premise…


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2


7. ~a (c d) by SPEC on 6

8. …




12. (~c (a b)) (~c (~a e))) by CONJ on 10, 11

13. (~c ((a b) (~a e))) Lemma 1 on 12


66

Lemma 2: p (q r) q (p r)

Continuing with

that premise…


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2



8. ~a (~c e) by SPEC on 6

9. …




13. (~c (a b)) (~c (~a e))) by CONJ on 11, 12

14. (~c ((a b) (~a e))) Lemma 1 on 13


67

AHA!!


Continuing with

that premise…

We treated

connecting these

as its own problem

and came up with

Lemma 2!


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2




9. …



12. ~c (~a e) by Lemma 2 on 8

13. (~c (a b)) (~c (~a e))) by CONJ on 11, 12

14. (~c ((a b) (~a e))) Lemma 1 on 13


68


Lemma 2 lets us

connect these

directly!

Now what. Let’s

pause, remind

ourselves what

our (sub)goals

are, and look at

what we have.

10


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2




9. …




13. (~c (a b)) (~c (~a e))) by CONJ on 11, 12

14. (~c ((a b) (~a e))) Lemma 1 on 13


69

Hmm..


How do we do

something with

this? Again, we

treated this as a

separate

problem:


Subproblem:

1. a b premise

2. …


70

Now we do our

usual. Get rid of

, work

backward, work

forward…

This time, we’ll

show you what we

did. We broke out

the goal and

starting point and

turned them into a

whole other proof

problem!


Subproblem:

1. a b premise

2. ~a b by IMP on 1

3. …

4. c ~a b “subgoal”

5. c (a b) by IMP on 4

6. ~c (a b) by IMP on 5

71

That’s about as far as dumping can take us.

But, look at step 2 and step 4. What’s the difference?


Subproblem:

1. a b premise

2. ~a b by IMP on 1

3. c ~a b by GEN on 2

4. c (a b) by IMP on 3


72

Great! We can always OR on something else.

We did it!

Let’s patch it back into the original proof.

But… could we have done it more easily? Question your solutions!

(Hint: check out line 4. How can you get there?)


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2




9. …


11. c (a b) “subgoal”



14. (~c (a b)) (~c (~a e))) by CONJ on 12, 13

15. (~c ((a b) (~a e))) Lemma 1 on 14


73

Patching in “step

4” of the

previous proof.

Can it get us

back to step 4 of

this proof?


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2




9. …


11. c (a b) by GEN on 4



14. (~c (a b)) (~c (~a e))) by CONJ on 12, 13

15. (~c ((a b) (~a e))) Lemma 1 on 14


74

Sure! In one

step!

Now what? Only

one subgoal left.

How does it

connect to the

top of the proof?

11


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2




9. …





14. (~c (a b)) (~c (~a e))) by CONJ on 12, 13

15. (~c ((a b) (~a e))) Lemma 1 on 14


75

Hmm…

That works if a is

false.

Can we make a

false?

What if a is true?


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. a b by SPEC on 2

5. ~a ((c d) (~c e)) by SPEC on 2




9. …





14. (~c (a b)) (~c (~a e))) by CONJ on 12, 13

15. (~c ((a b) (~a e))) Lemma 1 on 14


76

If a is true,

then c isn’t.

If c’s not true,

then c d is true.

Let’s put that in

logic!

I looked around for a

way to establish ~a

but couldn’t. So, I

checked what

happens if a is true.


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1

4. ~a ~c d by GEN on 3

5. a b by SPEC on 2

6. ~a ((c d) (~c e)) by SPEC on 2




10. …





15. (~c (a b)) (~c (~a e))) by CONJ on 13, 14

16. (~c ((a b) (~a e))) Lemma 1 on 15


77

We need to

“fabricate” a d.

The rest will be

just IMP

applications.


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1


5. ~a (c d) by IMP on 4

6. a (c d) by IMP on 5

7. a b by SPEC on 2

8. ~a ((c d) (~c e)) by SPEC on 2




12. …





17. (~c (a b)) (~c (~a e))) by CONJ on 15, 16

18. (~c ((a b) (~a e))) Lemma 1 on 17


78

Now, we put

these together,

and we’re done!


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

3. ~a ~c by DM on 1


5. ~a (c d) by IMP on 4

6. a (c d) by IMP on 5

7. a b by SPEC on 2

8. ~a ((c d) (~c e)) by SPEC on 2



11. (~a a) (c d) by CASE on 10, 6

12. T (c d) by NEG on 11

13. (c d) by M.PON on 12, T





18. (~c (a b)) (~c (~a e))) by CONJ on 16, 17

19. (~c ((a b) (~a e))) Lemma 1 on 18


79

(At step 13, no need

to separately

establish T. T is a

“tautology”; it’s

always true!)

QED!! Whew!


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

… …


80

So, what did that prove?

Technically: that if the conditions on the cond branches are

mutually exclusive (cannot both be true at the same time) and if

the result of the original version was true, then the version with switched cond branches will also be true.

In fact, if you go back and think carefully about the proof, we can

conclude something much bigger without too much more work: “If two conditions on neighboring cond branches are mutually

exclusive (and have no ‘side effects’), we can switch those

branches without changing the meaning of the program.”

12


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

… …


81

For reference: fruitless directions I tried include changing a b to ~a

b, attempting to form the negation of c d, and a bunch of other false

starts… all of which helped me build pieces I needed for my final

strategy!

You should have lots of scratchwork if you do a problem this large.


In prop logic:

1. ~(a c) premise

2. (a b) (~a ((c d) (~c e))) premise

… …


82

Exercise: For expressions a, b, and c that evaluate to Booleans (with no side effects), we can translate code like: (if a b c)

To logic like this instead of our usual: (a b) (~a c)

Prove that they’re equivalent.

Then, figure out how a cond would similarly translate.

Finally, go back and redo some of our proofs (like the one we just did)

with the new representation.

snick

snack

More problems to solve...

(on your own or if we have time)

83

Problem:

Who put the cat in the piano? Hercule Poirot has been asked by Lord Martin to find out who closed

the lid of his piano after dumping the cat inside. Poirot interrogates two of the servants, Akilna and Eiluj. One and only one of them put the cat in the piano. Plus, one always lies and one never lies.

Akilna says: – Eiluj did it.

– Urquhart paid her $50 to help him study.

Eiluj says: – I did not put the cat in the piano.

– Urquhart gave me less than $60 to help him study.

Problem: Whodunit?

84

Problem: Automating Proof

Given:

p q

p ~q r

(r ~p) s ~p

~r

Problem: What’s everything you can prove?

85

Problem: Canonical Form

A common form for propositional logic

expressions, called “disjunctive normal

form” or “sum of products form”, looks like

this:

(a ~b d) (~c) (~a ~d) (b c d

e) ...

In other words, each clause is built up of

simple propositions or their negations,

ANDed together, and all the clauses are

ORed together. 86

13

Problem: Canonical Form

Problem: Prove that any propositional logic

statement can be expressed in disjunctive

normal form.

87

Mystery #1

Theorem:

p q

q (r s)

~r (~t u)

p t

u

Is this argument valid or invalid? Is whatever u means true?

88

Mystery #2

Theorem:

p

p r

p (q ~r)

~q ~s

s

Is this argument valid or invalid? Is whatever s means true?

89

Mystery #3

Theorem:

q

p m

q (r m)

m q

p

Is this argument valid or invalid? Is whatever p means true?

90

Practice Problem (for you!)

Prove (with truth tables) that hypothetical

syllogism is a valid rule of inference:

p q

q r

p r

91


Prove (with truth tables) whether this is a

valid rule of inference:

q

p q

p

92

14


Are the following arguments valid?

This apple is green.

If an apple is green, it is sour.

This apple is sour.

Sam is not barking.

If Sam is barking, then Sam is a dog.

Sam is not a dog.

93


Are the following arguments valid?

This shirt is comfortable.

If a shirt is comfortable, it’s chartreuse.

This shirt is chartreuse.

It’s not cold.

If it’s January, it’s cold.

It’s not January.

Is valid (as a term) the same as true or correct (as English ideas)? 94

More Practice

Meghan is rich.

If Meghan is rich, she will pay your tuition.

Meghan will pay your tuition.

Is this argument valid?

Should you bother sending in a check for your

tuition, or is Meghan going to do it? 95

Problem:

Equivalent Java Programs Problem: How many valid Java programs

are there that do exactly the same thing?

96

Resources: Statements

From the Java language

specification, a

standard statement is

one that can be:

http://java.sun.com/docs/books/jls/third_edition/html/statements.html#14.5 97

Resources: Statements

From the Java language

specification, a

standard statement is

one that can be:


15

What’s a “Block”?

Back to the Java Language Specification:


What’s a “Block”?

A block is a sequence of statements, local class declarations and local variable declaration statements within braces.

…

A block is executed by executing each of the local variable declaration statements and other statements in order from first to last (left to right).

100

What’s an “EmptyStatement”

Back to the Java Language Specification:


Problem: Validity of Arguments

Problem: If an argument is valid, does that

mean its conclusion is true? If an

argument is invalid, does that mean its

conclusion is false?

102

Problem: Proofs and

Contradiction Problem: Imagine I assume premises x, y,

and z and prove F. What can I conclude

(besides “false is true if x, y, and z are

true”)?

103

Proof Critique

Theorem: √2 is irrational Proof: Assume √2 is rational, then...

There’s some integers p and q such that √2 = p/q, and

p and q share no factors.

2 = (p/q)2 = p2/q2 and p2 = 2q2

p2 is divisible by 2; so p is divisible by 2.

There’s some integer k such that p = 2k.

q2 = p2/2 = (2k)2/2 = 2k2; so q2 and q are divisible by 2.

p and q do share the factor 2, a contradiction!

√2 is irrational. QED

104

16

Problem: Comparing Deduction

and Equivalence Rules Problem: How are logical equivalence rules

and deduction rules similar and different,

in form, function, and the means by which

we establish their truth?

105

Problem: Evens and Integers

Problem: Which are there more of, (a)

positive even integers, (b) positive

integers, or (c) neither?

106

Learning Goals: Pre-Class Learning Goals: In-Classcs-121/2013W2/... · • Prelude: What Is Proof?...

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