Mcs lecture19.methods ofproof(1)

Mathematics for Computer Science: Logic and Discrete Structures 2013-14

Dr Konrad Dabrowski

Methods of Proof

1. Mathematical proofs. 2

2. Direct proofs. 4

3. Proof by contraposition. 6

4. Proof by contradiction. 10

5. Proof of equivalence. 15

6. Proof by cases. 17

7. Existence proofs. 19

8. Proof by counter-example. 21

9. Applications. 22

10. Induction. 29

Mathematical proofs

• A mathematical proof is a valid argument that establishes the truth of a mathematical statement.

• A proof is generally constructed using:– the hypotheses of the theorem (that is, the things assumed to be true in the

theorem that is to be proved)

– axioms known to be true

– previously proved theorems

– rules of inference (that is, allowable rules that can be used to infer new mathematical statements from existing ones).

• Up until now, we have been considering formal proofs (in logics).

• Now we focus on informal proofs, as applied by humans.

Mathematical Proofs: Slide 2 of 36


Theorems

• Theorems are usually stated using an informal version of first-order logic and often have one of the following structures:– “Every such-and-such is a such-and-such.”

– “There is a such-and-such that is not a such-and-such.”

– “If something is a such-and-such then it is a such-and-such.”

• Consequently, in order to prove many theorems we utilize our knowledge of first-order logic.

• Different proof methods are available to us, often depending upon the structure of the theorem to be proved.


Direct proofs

• A direct proof is used to prove theorems of the form:– “If such-and-such then such-and-such.”

• More generally, theorems of the form: x(P(x) Q(x)).

• Essentially, a direct proof is a list of statements starting from P(x) and ending at Q(x), and where every statement in the list is:– an axiom

– a previously proved theorem

– an inference of such using a rule of inference.

• Direct proofs, once stated, tend to be easy to check, but often some insight is required to devise the proof in the first place.

• (Compare this with Resolution, for example, where often there is a short sequence of resolutions yielding the required result but where it is not obvious as which resolutions to apply!)

Here is a direct proof of the following theorem:– “If n is an odd integer then so is n2.”

Proof Suppose that n is an odd integer.So, n = 2k + 1, for some integer k.Thus, n2 = (2k + 1)2 = (2k + 1)(2k + 1) = 4k2 + 4k + 1.When we divide n2 by 2 we get 2k2 + 2k remainder 1.Thus, n2 is odd. □

Here is a direct proof of the following theorem:– “If m and n are perfect squares then so is mn.”

Proof Suppose that m and n are perfect squares; so, m = a2, for some integer a, and n = b2, for some integer b.Thus mn = a2b2 = (ab)2, and so mn is a perfect square. □


Examples of direct proofs


Proof by contraposition

• Proofs by contraposition are generally used to prove theorems of the form: x(P(x) Q(x)).

• The key to a proof by contraposition is the following argument from first-order logic:– M is a model of x(P(x) Q(x))

iff for every x, M is a model of P(x) Q(x)

iff for every x, M is a model of Q(x) P(x)

iff M is a model of x(Q(x) P(x)).

• Thus, if we wish to prove that x(P(x) Q(x)) is a theorem then it suffices to prove that for any value of x, Q(x) P(x).


Examples of proofs by contraposition

Here is a proof by contraposition of the following theorem:– “If n is an integer and 3n + 2 is odd then n is odd.”

Proof Assume the negation of what we want to prove; that is, assume that n is even.So, n = 2k, for some integer k.Thus, 3n + 2 = 6k + 2 is even.Hence, if n is even then 3n + 2 is even; that is, if 3n + 2 is odd then n is odd. □

• Let’s try and prove the result above with a direct proof.Suppose that 3n + 2 is odd; so, 3n + 2 = 2k + 1, for some integer k.So, 3n = 2k – 1 is odd.Its not so clear as to how to proceed now, as we appear to be back where we started except we have that 3n is odd as opposed to 3n + 2.

Here is a proof by contraposition of the following theorem:– “If m and n are integers and mn is even then m is even or n is even.”

Proof Assume that m is odd and n is odd; so, m = 2k + 1, for some integer k, and n = 2p + 1, for some integer p.

Thus, mn = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(2kp + k + p) + 1 which is odd. □

Here is a proof by contraposition of the following theorem:– “If n is an integer and n3 + 5 is odd then n is even.”

Proof Assume that n is odd; so, n = 2k + 1, for some integer k.

Thus, n3 + 5 = (2k + 1)3 + 5 = (2k + 1)(4k2 + 4k + 1) + 5 = 8k3 + 12k2 + 6k + 6 = 2(4k3 + 6k2 + 3k + 3) which is even. □


Examples of proofs by contraposition

• The previous examples show that using one proof method can be easier than using another; but how do we decide which proof method to use?

• There is no canonical answer: probably the best approach is to try a direct proof and if you struggle then try a proof by contraposition!

Consider the following theorem:– “The sum of two rational numbers is rational.”

Proof Try a direct proof. Let x and y be rational numbers; so, let x = a/b and let y = c/d, where a, b, c, and d are integers.Thus, x + y = a/b + c/d = ad/bd + bc/bd = (ad + bc)/bd, which is rational, and so the direct proof works.If we tried a proof by contraposition then we would start by considering the irrational sum of two numbers x + y.This would not be of very much help! □


Which is the easiest method to use?


Proof by contradiction

• Suppose that we wish to prove that something, p say, is true.

Suppose also that we know– something else, q say, to be false p q to be true.

The only way that p q can be true when q is false is for p to be true.

The above state of affairs results in a proof by contradiction.

• Look more closely at a proof by contradiction.

It proves that p is true but it does it non-constructively; that is, it does it not by building a proof that p is true but by showing that if p were false then we would be able to prove that something known to be false is true!


Examples of proofs by contradiction

Here is a proof by contradiction of the following theorem:– “2 is irrational.”

Proof Suppose that 2 is rational; that is, suppose that 2 = a/b, where a and b are integers and where no integer apart from 1 divides both a and b.So, 2 = a2/b2, with a2 = 2b2 and a2 even.

Lemma: If a2 is even then a is even.

Proof of lemma [Proof by contraposition] Suppose that a is odd; that is, a = 2k + 1, for some integer k.Thus, a2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1; consequently, a2 is odd. □

Now let us continue with our main proof.


Examples of proofs by contradiction

Recall, we have that a2 = 2b2 and a2 is even.

So, by our lemma, we have that a is even; that is, a = 2p, for some integer p.

Hence, 4p2 = 2b2 with b2 = 2p2; in particular, b2 is even.

Applying our lemma again yields that b is even; so, b = 2q, for some integer q.

Thus, 2 divides a and 2 divides b, which yields a contradiction.

Hence, our original assumption that 2 is rational cannot be correct; that is, 2 is irrational. □

• Note: we have not actually constructed a description of the actual irrational number 2: we have simply shown that it cannot be rational and therefore must be irrational!

Consider the following theorem:– “If n is a positive integer and n is odd then 5n + 6 is odd.”

Proof [Direct proof] Let n be odd; so, n = 2k + 1, for some integer k.

Thus, 5n + 6 = 5(2k + 1) + 6 = 10k + 11 = 2(5k + 5) + 1, which is odd.

[Proof by contraposition] Suppose that 5n + 6 is even; so, 5n + 6 = 2k, for some integer k.

Thus, n + 4n = 2k – 6 and n = 2k – 6 – 4n = 2(k – 3 – 2n), which is even.

[Proof by contradiction] Let n be odd; that is, n = 2k + 1, for some integer k.

Suppose that 5n + 6 is even; that is, 5n + 6 = 2p, for some integer p.

Thus, 5(2k + 1) + 6 = 2p; that is, 10k + 5 + 6 = 2p; that is, 11 = 2(p – 5k).

Hence, 11 is even which yields a contradiction; thus, 5n + 6 is odd. □


More examples


More examples

• Note the structure of our proof by contradiction in the previous example:– we wanted to prove p q, so we rewrote this as p q and assumed its

negation, namely p q, to obtain our contradiction.

Here is a proof by contradiction of the following theorem:– “There is no rational number r for which r3 + r + 1 = 0.”

Proof Suppose that r = a/b is rational, where a and b are integers having no common factor different from 1.So, a3/b3 + a/b + 1 = 0 with a3 + ab2 + b3 = 0.As 0 is even, a3 + ab2 + b3 is even.If a and b are odd then a3 + ab2 + b3 is odd – contradiction.If a is odd and b is even then a3 + ab2 + b3 is odd – contradiction.If a is even and b is odd then a3 + ab2 + b3 is odd – contradiction.So, a and b are even and have a common factor 2 – contradiction. □


Proof of equivalence

• Proofs of equivalence have the form p q and are usually structured into two parts:– p q and q p.

Consider the following theorem:– “For any integer n, n is odd if, and only if, n2 is odd.”

Proof [Direct proof] If n is odd then n = 2k + 1, for some integer k.

So, n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 is odd.

[Proof by contraposition] Conversely, now we wish to prove that if n2 is odd then n is odd.

Suppose n is even; that is, n = 2k, for some integer k.

Thus, n2 = (2k)2 = 2(2k2) is even. □


Examples of proofs by equivalence

Consider the following theorem:– “If x is a real number then the following are equivalent:

i. x is rationalii. x/2 is rationaliii. 3x – 1 is rational.”

Proof Suppose that i. holds; so, x = a/b is rational and x/2 = a/2b is rational. So, ii. holds.Suppose that ii. holds; so, x/2 = a/b is rational and x = 2a/b is rational. So, i. holds.Suppose that i. holds; so, x = a/b is rational and 3x – 1 = 3a/b – 1 = (3a – b)/b is rational. So, iii. holds.Suppose that iii. holds; so, 3x – 1 = a/b is rational and x = (a + b)/3b is rational. So, i. holds.Thus, i. ii. and i. iii.Hence, ii. i. and i. iii.; and iii. i. and i. ii. So, ii. iii. □


Proof by cases

• Sometimes the proof of a theorem can be split up into the separate proofs of a small number of different cases.Consider the following theorem:– “For integers a, b, and c, min{a, min{b, c}} = min{min{a, b}, c}, where

min{x, y} is the minimum of x and y.”

Proof [Proof by cases] Case i. a b, c. min{a, min{b, c}} = a and min{min{a, b}, c} = min{a, c} = a. Thus, min{a, min{b, c}} = min{min{a, b}, c}.Case ii. b a, c. min{a, min{b, c}} = min{a, b} = b and min{min{a, b}, c} = min{b, c} = b. Thus, min{a, min{b, c}} = min{min{a, b}, c}.Case iii. c a, b. min{a, min{b, c}} = min{a, c} = c and min{min{a, b}, c} = c. Thus, min{a, min{b, c}} = min{min{a, b}, c}. □


Exhaustive checks

• Sometimes a proof by cases is nothing other than an exhaustive check.

Consider the following theorem:– “There are no positive cubes less than 1000 that are the sum of the cubes

of two distinct positive integers.”

Proof [Proof by contradiction] Suppose that the theorem is false; so, a cube x3 involved in the sum must be such that x3 < 1000.

Thus, x < 10 (as 103 = 1000) and the cubes involved in the sum come from the set {1, 8, 27, 64, 125, 216, 343, 512, 729}.

But no pair of numbers from this set are such that their sum is equal to another cube from this set – contradiction. □


Existence proofs

• Existence proofs tend to be proofs of theorems of the form xP(x).• In order to prove a theorem of the form xP(x) we can either construct

an actual x such that P(x) holds or we can show that such an x must exist without actually constructing it.

• The former are called constructive proofs; the latter non-constructive proofs.

For example, in order to prove the theorem:– “There is a positive integer that is the sum of all positive integers

less than it.”

we simply write 3 = 1 + 2 to constructively prove this.

On the other hand, consider the following theorem:– “There exist irrational numbers x and y such that xy is rational.”

Proof [Non-constructive existence proof] 2 is irrational (we proved this earlier).Define z = 22.Suppose that z is irrational.So, z2 = (22)2 = (2)22 = (2)2 = 2 is rational.Hence:– if z is irrational then z2 is rational– if z is rational then 22 is rational.

Either way, we have our result. □

• Of course, we haven’t actually constructed x and y as in the statement of the theorem; we’ve just shown that they exist!


Non-constructive existence proofs


Proof by counter-example.

• Consider a statement of the form xP(x).• In order to prove it, we need to prove that for every x, P(x) holds.• In order to disprove it, we need to find some x such that P(x) holds

(or, at least, show that such an x exists); that is, find (the existence of) a counter-example.

For example, consider the statement:– “If a and b are rational numbers then ab is rational.”

Refutation Put a = 2 and b = ½. Then ab = 2 is irrational. □

Hence, the statement is not a theorem as we have found a counter-example.


Applying our proof techniques

• A tiling of a chess-board (of an as yet undefined shape) is a placement of black-and-white dominoes so that every domino straddles two squares and so that the board is covered.

• Pictured are two tilings of an 8 8 chess-board and a 10 10 chess-board.

• It is trivial to prove that any x x chess-board, where x is even, can be tiled (how might you do this?).

• However, suppose that we have an x x chess-board with x odd: can it be tiled?

No; as such a chess-board has an odd number of squares and any number of dominoes covers an even number of squares [proof by contradiction].



• Suppose that our chess-board is a mutilated 4 4 chess-board with a pair of opposite squares removed.

• Can such a chess-board be tiled?• We shall try and prove that the

answer is “no”, and we shall try and obtain a proof by contradiction.

• First, we may assume, without loss of generality, that it is the top-left and bottom-right pair of squares that has been removed.

• Suppose otherwise.• A simple rotation reduces us to the

situation where it is the top-left and bottom-right squares that have been removed.



• [Proof by contradiction] Suppose that a tiling of our chess-board exists.

• [Proof by cases]• Case i. A domino covers squares 2 and 3.

This forces the remaining dominoes to be placed deterministically.

So, squares 13 and 15 cannot be tiled.• Case ii. A domino covers squares 2 and 6.

This forces the remaining dominoes to be placed deterministically.

So, square 15 cannot be tiled.

This rules out all possibilities.

Hence, by splitting the proof into 2 cases and obtaining contradictions, we establish that the mutilated chess-board cannot be tiled. □

2 3 4

6 7 8

13 14 15

10 11 12

5

9


Another proof

• We can obtain another proof that our mutilated chessboard cannot by tiled with dominoes.

• [Proof by contradiction]

• Note that in any tiling of a chessboard (mutilated or otherwise), every domino covers a black and a white square.

Thus, if a tiling exists then the number of white squares must be equal to the number of black squares.

In our mutilated chessboard this is not the case; hence, we obtain a contradiction.

• If there are black squares and white squares missing then this argument doesn’t suffice.

2 3 4

6 7 8

13 14 15

10 11 12

5

9



• Let’s try and prove something about tilings of bigger chess-boards.• Consider the following statement:

– “When a black square and a white square are removed from an 8 8 chess-board, the remaining chess-board can always be tiled.”

Here, we are allowing the removal of any black square and any white square.

• We shall prove that this statement is true.

The statement is of the form xP(x), where x ranges over all possible placement of tiles (of any back-white-square-mutilated chessboard) and P(x) holds if, and only if, x is a tiling of the chess-board.

We shall provide a constructive existence proof.



• Number the squares as shown and note that we can lie dominoes nose-to-tail along the path shown (assuming all squares are present).

• Note also that squares on the chess-board alternate between white and black along this path.

2 3 4

131415

1 6 7 8

91112

5

1064

181716 2019 2163 22

272829 2526 2462 23

323130 3433 3561 36

414243 3940 3860 37

464544 4847 4959 50

555657 5354 5258 51


20

57


• Remove any black square b and any white square w.

• Note that the number of squares between b and w along the path is even.

Also, the number of squares along the path from w to b is even too.

• Thus we can lay dominoes nose-to-tail and cover all the squares of the mutilated chess-board.

• The result follows. □• How would you generalize

the proof to an x x chess-board, where x is even?

2 3 4

131415

1 6 7 8

91112

5

1064

181716 19 2163 22

272829 2526 2462 23

323130 3433 3561 36

414243 3940 3860 37

464544 4847 4959 50

5556 5354 5258 51


Induction

• Possibly the most important proof technique in Computer Science is induction.

• Induction is a technique which allows us to prove statements of the form nP(n), where n is a positive integer and P is some predicate depending upon n.

• Induction allows us to make the following inference:– “If we can prove:

• P(1) n(P(n) P(n + 1))

then we can infer nP(n).”

• It is important to remember that induction is a proof technique and not a technique for inferring theorems in some logical proof system.


Induction

• Proofs by induction come in two parts:– the base case, that is, P(1)

– the inductive step, that is, proving that if P(n) holds then P(n + 1) holds, where n is an arbitrary integer.

• The crucial point about the inductive step is that the chosen n is an arbitrary integer; that is, it can be any integer at all.

• Induction can be thought of as climbing an infinite ladder:– the base case equates to “I can stand on the bottom rung.”

– the inductive step equates to “whenever I am standing on some rung of the ladder then I can climb to the next rung.”

If both these statements are true then induction tells us we can climb the infinite ladder!


Examples of proofs by induction

• When tackling proofs by induction, perhaps the most important and difficult thing to do is to formulate the induction hypothesis; that is, P(n).

Consider the following theorem:– “For any positive integer n, 1 + 2 + … + n = n(n + 1)/2.”

Proof [Proof by induction] Our induction hypothesis, P(n), is:– “1 + 2 + … + n = n(n + 1)/2”.

Note that as to whether P(n) holds might depend upon the value of n.

Base case: put n = 1 in the inductive hypothesis and see whether it holds:– left-hand side is 1– right-hand side is 1(1 + 1)/2 = 1.

So, P(1) holds.



Inductive step: suppose that P(n) holds; that is, – 1 + 2 + … + n = n(n + 1)/2.

Consider P(n + 1).In order to decide what P(n + 1) states, simply replace the symbol n in P(n) with n + 1; so, we obtain:– 1 + 2 + … + (n + 1) = (n + 1)((n + 1) + 1)/2.

We need to show that P(n + 1) holds if P(n) holds.Take the left-hand side of P(n + 1), namely:– 1 + 2 + … + (n + 1) = 1 + 2 + … + n + (n + 1)

= n(n + 1)/2 + (n + 1) (since P(n) holds)= (n + 1)(n + 2)/2.

But this is identical to the right-hand side of P(n + 1), and so P(n + 1) holds if P(n) does.Thus, by induction, P(n) holds for every positive integer n. □



• We shall prove the following theorem using induction:– “Any rectangular chess-board with an even number of squares can be tiled

with dominoes.”

Proof Let our induction hypothesis P(n) be:– “For every m < n, either m is odd, or m is even and any rectangular chess-

board with m squares can be tiled with dominoes.”

Base case: P(1) states:– “for every m < 1, either m is odd, or m is even and any rectangular chess-

board with m squares can be tiled with dominoes”.

But this is vacuously true as there is no m < 1 making the statement false.

(Note: it is quite often the case that base cases of inductions are trivial.)



Inductive step: So, let us suppose that P(n) holds, for some n; that is:– “For every m < n, either m is odd, or m is even and any rectangular chess-

board with m squares can be tiled with dominoes.”

Consider P(n + 1).If n + 1 is even then n is odd and so combined with P(n), above, we must have that:– “For every m < n + 1, either m is odd, or m is even and any rectangular

chess-board with m squares can be tiled with dominoes.”

So, the interesting situation is when n + 1 is odd, i.e., when n is even.

Consider some rectangular chess-board with n squares. If we can prove that it can be tiled with dominoes then we are done.As our rectangular chess-board has an even number n of squares then either the number of rows is even or the number of columns is even.Without loss of generality, suppose that the number of rows r is even.



Consider the chess-board with the squares in the first column removed.

As the number of rows r is even, the amended chess-board has n – r < n squares, which is even.

By the induction hypothesis, this amended chess-board can be tiled.

However, this tiling can be extended to the original chess-board by placing a column of dominoes in the first column of the chess-board.

The result follows by induction. □

amended chess-board, tiled by induction

tiled first column


Summary

• Much of mathematical reasoning is based upon propositional logic and first-order logic.

• In day-to-day life we use reasoning based upon these logics all the time without even realising it.

• The ability to reason is crucial and fundamental to Computer Science, especially the capacity to automate (mathematical) reasoning.

• Of particular relevance to Computer Science is the constructive nature of reasoning, for in Computer Science much of our activity is taken up with building things (such as programs or proofs or data structures or arguments or …).

• Perhaps the most striking relationship between proofs and programs is the intimate relationship between induction and recursion, which share the same intuition: build larger artifacts out of previously-built smaller ones.

Technology

Mcs lecture19.methods ofproof(1)