This is the resource page for Section 2 of Logic and Computation, CSU 290 (Spring 2008). Welcome.
The purpose of this page is to simply complete the lectures, partly to keep us in sync with the other section, partly because there is only so much time and much interesting things to discuss.
These notes are meant to supplement the lectures, the readings and the lecture notes (available on the course web site) and not replace them.
Feel free to comment at the bottom of the page if you have any questions or anything is unclear. (Select "Add Comment" under "Add Content" down below.) I will generally merge your comments and questions into the main text when I get to them.
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February 15: Formal proof for
Here is a complete formal proof for
~(t=nil) using the axioms I gave in class. (This is the proof that I started on Monday, but ran out of time 5 steps into the proof.)
You should be able to understand all the steps in that proof, if not be able to come up with the proof yourself. We will not ask you to come up with these kind of formal proofs in this course. Instead, we will use the more informal "chain of equals" or "chain of relations" approach I presented in class.
In the proof below, I write
~A for "not
A & B for "
Okay, so 2 is something we will use later on. Let's remember it.
Excellent. So now we can see that 2 and 11 together essentially give us that
t /= nil.
We just have to work some to get that to come out of the proof.
And we are just about done. We have both antecedents of the implication in 16.
February 20: Review of Proving Theorems
Here is what we have seen until now as far as proving theorems are concerned.
We reviewed Boolean logic and the concept of a tautology of Boolean logic. Recall that a tautology is just a Boolean formula that is always true, irrespectively of the truth values you give to the Boolean variables in the formula.
We shall take Boolean (aka propositional) reasoning for granted.
We introduced the ACL2 logic, as essentially Boolean logic augmented with the ability to reason about basic formulas of the form EXP1 = EXP2, where EXP1 and EXP2 are ACL2 expressions.
The ACL2 logic is defined formally using the following axioms and rules of inferences:
- Every tautology of Boolean logic is an axiom
x = x(reflexivity)
x = y => y = x(symmetry)
x = y /\ y = z => x = z(transitivity)
x1 = y1 /\ ... /\ xn = yn => (f x1 ... xn) = (f y1 ... yn)for every function symbol
fof arity n
Fis an arbitrary formula, and s is a substitution of ACL2 expressions for free variables in
F => Gderive
We also have axioms corresponding to the different primitives of the language. The basic ones are:
x = y => (equal x y) = t
x /= y => (equal x y) = nil(where
/=represents "not equal to")
x = nil => (if x y z) = z
x /= nil => (if x y z) = x
(car (cons x y)) = x
(cdr (cons x y)) = y
consp (cons x y) = t
consp (nil) = nil
(integerp x) = t => (consp x)=nil
Additionally, we can take as axioms the basic properties of arithmetic that you have seen in high school and discrete maths. Thus, when proving theorems, you will be free to perform standard simplification of arithmetic.
A formal proof of
F is a sequence of formulas such that every formula in the sequence is either an axiom or derived from previous formulas using an inference rule, and such that the last formula of the sequence is
I gave you a formal proof of
t /= nil above. We shall not write a lot of formal proof, and you certainly will not have to write any of them.
Instead, we shall use some proof techniques to prove formulas that have a certain form. These techniques do not work for all formulas, but they work for enough of them to be interesting.
If the formula you want to prove is of the form EXP1 = EXP2, then one technique is to use a sequence of equalities to equate EXP1 and EXP2:
where all the equalities are justified using the axioms or the inference rules above. By transitivity of =, this clearly shows that EXP1 = EXP2.
Alternatively, it is sometimes easier to simplify both EXP1 and EXP2 into a common expression EXP3.
Again, it should be pretty clear that this proves that EXP1 = EXP2. (By transitivity of = once again.)
If the formula you want to prove is of the form HYPS => EXP1 = EXP2, where HYPS is some hypotheses (formulas), then you can use a variant of the above, that is, prove that EXP1 = EXP2 using a sequence of equalities (or simplify EXP1 and EXP2 to the same expression EXP3), except that when you derive the equalities you can in addition to the axioms and inference rules use the hypotheses in HYPS to help you.
February 25: Solutions to Exam 3 Proof Questions
Recall the definitions:
Question 2. Prove
(consp x) => (len (car (add99 x))) = (+ (len (car x)) 1)
Question 3. Prove
(endp x) => (same-lenp (add99 x) y)
Question 4. Prove
((same-lenp x y) /\ (consp x) /\ (same-lenp (add99 (cdr x)) (+ y 1))) => (same-lenp (add99 x) (+ y 1))
(You can use the fact that
((same-lenp x y) /\ (consp x)) => (+ 1 (len (car x))) = (+ 1 y).)
February 26: On Termination
Here is the argument that, at least in Scheme, it is impossible to write a function that correctly determines whether another function terminates. We argue by contradiction.
Suppose that we could write a function (terminates? f x) that returns #t (true) when (f x) terminates, and returns #f (false) when (f x) does not terminate. I don't care how terminates? is written - just suppose you managed to write it. I will not show that you get something completely absurd as a result.
In particular, I can now write the following perfectly legal scheme function:
My question: does (oops empty) terminate?
Let's see. It either does or doesn't. So let's consider both cases.
- If (oops empty) terminates, then by assumption (terminates? oops empty) must be true, and therefore, by the code of oops, (oops empty) must loop forever, i.e., not terminate, contradicting that (oops empty) terminates.
- if (oops empty) does not terminate, then by assumption (terminates? oops empty) must be false, and therefore by the code of oops, (oops empty) returns 1 immediately, i.e., terminates, contradicting that (oops empty) does not terminate.
Because we get a contradiction no matter what, it must be that what we initially assumed was wrong, i.e., that terminates? works correctly. In other words, we cannot write a correct terminates? function in Scheme.
It turns out that this argument can be made to work for any programming language, but you'll have to wait for CSU 390 to see it.
One approach to arguing that a recursive function terminates is to argue that we are making progress towards the base case. The design recipe that we gave you and that you learned in 211 is in fact meant to ensure exactly this, at least for the kind of functions that we have seen until now.
It is sometimes a little subtle to argue that you are indeed making progress towards termination. In particular, consider the following ACL2 function:
(Assume that we have correct definitions for
oddp.) I claim that this terminates for all inputs. Clearly, if the input is not a natural number, it terminates immediately. (Why?) Otherwise, can we argue that we are making progress towards the base cases (either 0 or 1)? Here are two arguments: the first, the one raised in class on Monday, is that that even though when the input is odd the recursive call is done on a bigger number (that therefore does not immediately progresses towards the base case), (foo (+ n 1)) will be called on an even number, which means that at the following iteration, foo will be invoked on (/ (+ n 1) 2), which is smaller than n. Thus, even though at every step we are not making progress towards the base case, we are making progress towards the base case at either every step or every two steps, depending on whether the input is even or odd. That's actually sufficient to establish termination. (Why?)
Here's another way of thinking about it that actually shows progress towards termination at every step. Write the input as a binary number, and consider the following number <n> derived from the input n:
- 4 is 100 in binary, so <4> = 1 + (2 * 3) - 2 = 1 + 6 - 2 = 5,
- 5 is 101 in binary, so <5> = 2 + (2 * 3) - 0 = 2 + 6 = 8,
- 6 is 110 in binary, so <6> = 2 + (2 * 3) - 1 = 2 + 6 - 1 = 7.
You can check that at every recursive step of the function, foo is called on an input n whose <n> is strictly smaller than the original input to the function. And we are making progress towards the base case, with <1> = 3.
However, this doesn't work for every function, even those that look similar to the above. One of the most famous examples is the following so-called Collatz function:
It is unknown whether this function terminates on all inputs. Plot out a few calls for small values of n, and you'll get a sense for how long some of the iterations take. Termination has been checked for all natural numbers up to 10^16, but for all we know it fails to terminate on some input greater than 10^17. We just don't know enough about number theory to say anything else.
I posted some more mathematical references to this function here.
February 27: The Final Bit of the Proof that
(len (add1 a)) = (len a)
I went over it quickly at the end of lecture today, so it pays to look at it more slowly.
Recall the definitions I gave in class:
I claimed that in order to prove that
(len (add1 a)) = (len a), it suffices to prove the following two theorems:
We saw that these are actually quite easy to prove. (Do it again if you're shaky on the details.)
We need to argue, though, that
Add1ConsP together give you that
(len (add1 a)) = (len a) for all possible
a in the ACL2 universe.
We are going to argue this by contradiction. So assume that it is not the case that
(len (add1 a)) = (len a) for all
Let A be the set of all values in the ACL2 universe for which
((len (add1 a)) is not equal to
(len a). By assumption, A is not empty.
s be an element of A that has smallest length amongst all elements of A. (If more than one element has smallest length, pick any of them.)
First off, we know that
s cannot be an atom. Why? Because we proved
Add1EndP, which says that atoms cannot be in A. So
s must be a cons pair, that is,
(consp s) = t.
s is a cons pair,
(cdr s) must have length strictly less than that of
s was chosen to have smallest length in A, then
(cdr s) cannot be in A.
(cdr s) is not in A, then it must be that
(len (add1 (cdr s))) = (len (cdr s)). (Recall how we defined A in the first place.)
So we know that
(consp s) = t, and that
(len (add1 (cdr s))) = (len (cdr s)). But we proved
Add1Consp, which says that it must be the case that
(len (add1 s)) = (len s). And this contradicts the fact that
s is in A in the first place!
We derive a contradiction, so what we assumed in the first place must be false - so there cannot be any
a such that
(len (add1 a)) is not equal to
(len a), that is,
(len (add1 a)) = (len a) for all
a in the universe.
March 11: Some Easy Proof Exercises
Consider the following definitions:
Make sure you understand them. You can assume the following properties about
EDIT: Let me add the following properties that you can assume too:
(You can give a shot at proving these, but the proof is a bit different than what we're used to - it requires case analysis, that you can do using only Boolean reasoning. For now, just take the properties above as true. I will talk about the proof technique for this in some other entry below.)
Now, let's write a function that applies
andp to a list of values, so that it returns true exactly when not all values in the list are nil.
Note what happens when we evaluate
(and-foldp '()), or when we evaluate
(and-foldp 5), or when we evaluate
Let's do something similar for
Again, note what happens when we evaluate
(or-foldp '()), or when we evaluate
(or-foldp 5), or when we evaluate
Let's prove some theorems. (Many of these are proof obligations obtained from the induction principle - can you identify the formula that these proof obligations end up proving?) Make sure that you understand what the theorem you are trying to prove is actually saying! Otherwise, this exercise is just meaningless symbol pushing, which is going to be excruciatingly boring.
(endp x) => (booleanp (and-foldp x))
(consp x) & (booleanp (and-foldp (cdr x))) => (booleanp (and-foldp x))
(endp x) => (booleanp (or-foldp x))
(consp x) & (booleanp (or-foldp (cdr x))) => (booleanp (or-foldp x))
(and-foldp (cons a b)) = (andp a (and-foldp b))
(or-foldp (cons a b)) = (orp a (or-foldp b))
Prove the following theorems, using the standard definition of
(endp x) => (and-foldp (app x y)) = (andp (and-foldp x) (and-foldp y))
(consp x) & (and-foldp (app (cdr x) y)) = (andp (and-foldp (cdr x)) (and-foldp y)) => (and-foldp (app x y)) = (andp (and-foldp x) (and-foldp y))
(endp x) => (or-foldp (app x y)) = (orp (or-foldp x) (or-foldp y))
(consp x) & (or-foldp (app (cdr x) y)) = (orp (or-foldp (cdr x)) (or-foldp y)) => (or-foldp (app x y)) = (orp (or-foldp x) (or-foldp y))
EDIT: note that proving the above gives you that the following two theorems are true:
(and-foldp (app x y)) = (andp (and-foldp x) (and-foldp y))
(or-foldp (app x y)) = (orp (of-foldp x) (or-foldp y))
These may come in handy in the proofs of the next theorems.
For the next few theorems, you will want to prove the following lemma first:
(consp x) => (app (list (car x)) (cdr x)) = x
(list z) is just an abbreviation for
(cons z nil). You can use the following axiom for cons:
(consp x) => (cons (car x) (cdr x)) = x.)
Use the standard definition of
Prove the following theorems:
(endp x) => (and-foldp (rev x)) = (and-foldp x)
(consp x) & (and-foldp (rev (cdr x))) = (and-foldp (cdr x)) => (and-foldp (rev x)) = (and-foldp x)
(endp x) => (or-foldp (rev x)) = (or-foldp x))
(consp x) & (or-foldp (rev (cdr x))) = (or-foldp (cdr x)) => (or-foldp (rev x)) = (or-foldp x)
March 17: Proof of
(true-listp x) => (rev (rev x)) = x
Here is the proof of
(true-listp x) => (rev (rev x)) = x that I did in class, complete with all auxiliary lemmas.
We proceed by induction on
x, meaning that we have to prove the following proof obligations:
(endp x) & (true-listp x) => (rev (rev x)) = x
(consp x) & ((true-listp (cdr x)) => (rev (rev (cdr x))) = (cdr x)) & (true-listp x) => (rev (rev x)) = x
(endp x) & (true-listp x) => (rev (rev x)) = x:
This proof relies on the following lemma, which says that if we have a true list and it is an atom, then it must be
nil. Note that we use this lemma to add a new result to the context, that is, to the set of known facts that include the hypotheses.
Proof of Lemma 1:
(endp x) & (true-listp x) => x = nil
Back to the main proof, where we now have to prove the inductive case:
(consp x) & ((true-listp (cdr x)) => (rev (rev (cdr x))) = (cdr x)) & (true-listp x) => (rev (rev x)) = x:
The proof uses two lemmas. Let's take care of the third lemma first.
Proof of Lemma 3:
(true-listp x) & (consp x) => (true-listp (cdr x)):
That was easy. Slightly harder is lemma 2, the one that Eric proved on the board today.
Proof of Lemma 2:
(rev (app a b)) = (app (rev b) (rev a)):
All that's missing now is Lemma 4, that says that
(rev b) is a true list for any
b. I left it as an exercise for you to prove for next time.
April 9: Topics for Exam #6
Topics covered include everything we have learned about the ACL2s theorem prover, lectures 25 to 29 on the web page; of course, you should not forget the basics of proving theorems.
Here are the main things we will test.
- Understanding ACL2s proof attempts
That was the largest component of these last two weeks.
In particular, understand the waterfall model (cf lecture 25), and be able to extract lemmas to be proved when ACL2s is stuck on a proof.
- Understand rewriting
Cf lecture 26 and the textbook, and see below.
- Understand decision procedures for propositional reasoning and congruence closure
Cf lecture 29 and see below.
Recall that when you prove a theorem (possibly with hypotheses) with a conclusion of the form LHS = RHS, this defines a rewrite rule that rewrites occurrences of LHS into RHS. (Note the direction!) Variables in the LHS are taken to be placeholders, that can be substituted for when rewriting happens.
The sequencing of rewrites when rewriting happens is important. The two main rules are:
- Rewriting occurs for inner expressions before outer expressions. Thus, if you have an expression
(f (g x)), the rewriter will attempt to rewrite
(g x)first, and then attempt to rewrite
- Rewriting occurs with the latest rewrite rules defined tried first, all to way to the oldest rewrite rules. In other words, if two (or more) rewrite rules are applicable to a given expression, the most recently admitted rewrite rule is used.
Decision Procedure for Propositional Reasoning
How does ACL2s decided whether a Boolean expression is a tautology or not? It could build a truth table, but it doesn't. (Well, actually, it almost does - it just uses a more efficient representation than a truth table, that contains essentially the same information.) Here's how it does it. For simplicity, here, I will only consider the case where there are no occurrence of functions others than
or in the Boolean expression. (It's an interesting exercise to figure out what happens when you do allow other functions.)
First, it rewrites the Boolean expression so that it does not contain any
not. It does so by replacing every occurrence of those functions by a corresponding
if expression. Here are the replacements:
(Convince yourself that these are correct by trying out a few examples, that is, seeing whether
(or t nil) evaluates to
t when replaced by the corresponding
if expression, etc.)
Thus, for example:
Once that is done, you are left with a bunch of
if expressions. For reasons that will become clear, we want to simplify the condition of those
if expressions as much as possible. In particular, we want all
if to have simple variables as conditions. To do so, we will apply the following transformation to an
if expression that has a non-variable as a condition. (If it's not a variable, it has to be an
if expression itself.)
(Convince yourself that you understand why applying this transformation does not change the value returned by the whole expression.)
Do this repeatedly until all
if expressions have a variable (or
nil) as a condition.
Continuing our example above:
Now, let's simplify things further.
First, take every
if expression, from the outermost to the innermost:
- if the expression is of the form
(if t X Y), replace it by
X, and repeat going into
- if the expression is of the form
(if nil X Y), replace it by
Y, and repeat going into
- if the expression is of the form
(if v X Y), where
vis a variable, replace every occurrence of
tand every occurrence of
nil, and repeat (going into
Continuing our example:
When you're done all of this, you have an expression with only
if expressions and with the property that for any "path" through the conditionals, you encounter every variable only once, in the condition of an
if. A Boolean expression is a tautology exactly when every "path" through the conditionals lead to a value that is guaranteed to be non-nil.
If a branch has a return value that is not guaranteed to be non-nil (for instance, it is
nil, or it is a variable), you can read off a counter example (that is, an assignment of truth values to variables that makes the Boolean expression false) by looking at the "path" through the conditionals that led to the values that is not guaranteed to be non-nil, setting a variable to
t if you ended up going to the THEN branch of the conditional, or to
nil if you ended up going to the right branch of the conditional.
Thus, in the above example, we ended up with
(if a nil nil); there are two "paths" into this conditional that lead to a value not guaranteed to be non-nil. One of those "path" is obtained by setting
t, and the other is obtained by setting
nil. It's easy to check that both of those assignments to
a are counterexamples to the initial expression being a tautology.
For a more complex example, consider
(implies (or (not a) b) (or a (not b))).
If you work through the above steps, you get the simplified form
(if a (if b t t) (if b nil t)). There is one "path" in the conditionals that lead to a value not guaranteed to be non-nil, and that's the "path" that leads to the
nil, which is obtained by setting
nil and setting
t. And you can check that
a=nil,b=t is a counterexample to the initial conjecture.
Decision Procedure for Equivalences