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

2. 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

3. 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.

4. Notation An intuitive way to write an argument with a set of hypotheses  and conclusion f is as follows:  ---  f hypotheses conclusion

5. Example proof 4 Show that: is a valid argument (p  q)  (p  r) r -------  q

6. 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))

7. 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)

8. 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

9. 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

10. Reductio ad Absurdum You’ve already met ‘proof by contradiction’ as a rule of deduction: This is also known as ‘Reductio ad Absurdum’  p  (r  r) ---------------  p

11. 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”

12. 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

13. Formalisation of assumptions The 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)

14. Formalisation of assumptions (continued) So,  = { (a  q )  m, ( t  q)  (  p  a) } These are the assumptions

15. 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  ?

16. Formal notation In our formal notation, the argument becomes: (a  q )  m ( t  q)  (  p  a) -------------------------   m  (  t  p )

17. Is this argument valid? 2 assumptions (a  q )  m ( t  q)  (  p  a) 1 conclusion  m  (  t  p ) 5 atomic propositions implies 2 5 = 32 different allocations of truth values to atomic propositions

18. 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)

19. Proof by contradiction is NOT “…where you prove that something is true by proving that it is false” Anon., EE2F1 exam 2002

20. Example Proof that is not a rational number

21. 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, then  m must be true, and (  t  p ) must be false

22. 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)

23. 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 true 10. But this contradicts the assertion that m is false in part 3 on the previous slide

24. 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

25. 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.

26. Identification of atomic propositions Atomic propositions b – the big bang theory is correct t – there was a time before anything existed w – the world will come to an end Formal statement of premises: b  (t  w) ww Formal statement of conclusion:  t   b

27. 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

28. Proof (continued) 1.If  t   b is false, then  t is true and  b is false Hence 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 true 4.But t is false. Hence w must be true. This contradicts assertion that  w is true 5.Hence  ⊨ f

29. 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

30. Adequacy A set of propositional connectives is adequate if For any set of atomic propositions p 1,…,p N and For any truth table for these propositions, There is a formula involving only the given connectives, which has the given truth table.

31. 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

32. Summary More analysis of arguments Proof by contradiction