EE1J2 – Discrete Maths Lecture 5 Analysis of arguments (continued) More example proofs Formalisation of arguments in natural language Proof by contradiction.

EE1J2 – Discrete Maths Lecture 5 Analysis of arguments (continued)

More example proofs Formalisation of arguments in natural language Proof by contradiction

Logical Consequence Let be a set of formulae and f a formula

f is a logical consequence of if for any assignment of truth values to atomic propositions for which all of the members of true, f is also true

If f is a logical consequence of , write ⊨f Note: this is consistent with ⊨f when f is a

tautology

Arguments

An argument consists of: A set of formulae, called the assumptions or

hypotheses A formula f, called the conclusion

If ⊨f then the argument is a valid argument In other words, an argument is valid if its

conclusion is a logical consequence of its assumptions.

Notation An intuitive way to write an argument with

a set of hypotheses and conclusion f is as follows:

---f

hypotheses

conclusion

Example proof 4 Show that:

is a valid argument

(p q)(p r)

r------- q

Proof 4(1) (p q)(2) (p r) (3) r

(4) p r (from (2))

(5) r p (from (4))(6) r p (from (5))

(7) p q (from (1))

(8) r q (from (6) and (7))

(9) q (from (8) and (3))

Alternative proof

Assume that the conclusion is false i.e q is False Therefore p must be true (from (1)) But p and r cannot both be true, by (2) Therefore r is false But this contradicts (3), so assumption must have been wrong

(p q)(p r)

r------- q

(1)(2)(3)

Proof by Contradiction This is an example of proof by

contradiction Basic idea is:

Assume that the conclusion is false Use this to deduce a contradiction Hence the conclusion must be true

Proof by Contradiction

Proof by contradiction is another powerful technique to show that an argument is valid

‘Proof by contradiction’ is also known as reductio ad absurdum

Reductio ad Absurdum You’ve already met ‘proof by contradiction’

as a rule of deduction:

This is also known as ‘Reductio ad Absurdum’

p(rr)---------------

p

Analysis of an Argument“The meeting can take place if all members are informed in advance, and it is quorate. It is quorate provided that there are at least 15 members present, and members will have been informed in advance if there is not a postal strike. Therefore, if the meeting was cancelled, there were fewer than 15 members present or there was a postal strike”

Identification of atomic propositions Atomic propositions are:m – the meeting takes place

a – all members have been informed in advance

t - there are at least 15 members present

q – the meeting is quorate

p – there is a postal strike

Formalisation of assumptionsThe meeting can take place if all members are informed in advance, and it is quorate

becomes (a q ) m

It is quorate provided that there are at least 15 members present, and members will have been informed in advance if there is not a postal strike

becomes ( t q) ( p a)

Formalisation of assumptions (continued) So,

= { (a q ) m , ( t q) ( p a) } These are the assumptions

Formalisation of conclusion

The argument concludes:

Therefore, if the meeting was cancelled, there were fewer than 15 members present or there was a postal strike

which becomes: m (t p )

So f = m (t p )

Is f a logical consequence of ?

Formal notation In our formal notation, the argument

becomes:

(a q ) m

( t q) ( p a)

-------------------------

m (t p )

Is this argument valid? 2 assumptions

(a q ) m ( t q) ( p a)

1 conclusion m (t p )

5 atomic propositions implies 25 = 32 different allocations of truth values to atomic propositions

Proof by Contradiction

Proof by contradiction Assume ⊨f is false Then there is an allocation of truth values to atomic

propositions for which all of the formulae in are true but f is false – called a counter-example

Show that the existence of a counter-example leads to a contradiction (e.g. that one of the formulae in must be false)

Proof by contradiction is NOT “…where you prove that something is true

by proving that it is false” Anon., EE2F1 exam 2002

Example Proof that is not a rational number2

Example Proof by Contradiction

1. Suppose there exists an assignment of truth values to m, a, t, q and p such that

(a q ) m, and ( t q) ( p a)

are both true, but

m (t p ) is false

2. If m (t p ) is false, thenm must be true, and (t p ) must be false

Proof continued…

3. It follows that

m is false, t is true and p is false

4. Now consider the first formula in , namely ( t q) ( p a)

5. Since this is true, t q and p a must both be true

6. Hence a and q are true, because t and p are true (from above)

Proof continued…

7. Finally consider the second formula in , namely (a q ) m

8. Since q is true and a is true (from 6 on the previous slide), a q is true,

9. Hence m must be true10. But this contradicts the assertion that

m is false in part 3 on the previous slide

Summary In summary, we have shown that the existence of

an assignment of truth values for which is true and f is false leads to a contradiction.

Hence such an assignment cannot exist. Hence ⊨f

Example 2

If the Big Bang theory is correct, then either there was a time before anything existed, or the world will come to an end. The world will not come to an end. Therefore, if there was no time before anything existed, the Big Bang theory is incorrect.

Identification of atomic propositions Atomic propositions

b – the big bang theory is correctt – there was a time before anything existedw – the world will come to an end

Formal statement of premises:b (t w)w

Formal statement of conclusion:t b

Proof by contradiction Formally, if

= {b (t w), w},

f is t b

Is it the case that ⊨ f ? Assume that f is not a logical consequence of Then there is an assignment of T and F to the

atomic propositions such that each formula in is true and f is false

Proof (continued)1. If t b is false, then

t is true and b is falseHence t is false and b is true

2. Now use the fact that, by assumption, b (t w) is true

3. Since b is true, (t w) must be true4. But t is false. Hence w must be true. This

contradicts assertion that w is true

5. Hence ⊨ f

Summary In summary, we have shown that the existence of

an assignment of truth values for which is true and f is false leads to a contradiction.

Hence such an assignment cannot exist. Hence ⊨f

Adequacy A set of propositional connectives is adequate if

For any set of atomic propositions p1,…,pN and For any truth table for these propositions, There is a formula involving only the given

connectives, which has the given truth table.

Adequacy The goal of the next lecture will be to show

that the set {, , , } is adequate and contains redundancy, in the sense that it contains subsets which are themselves adequate

We shall also introduce other sets of adequate connectives

Summary More analysis of arguments Proof by contradiction