- A step by step argument
- Should be verifiable and finite
- Every statement must be something we already know from the start such as a definition, axiom or previously-proved theorem
- Or a logical consequence of some conjunction of previous statements
- If you've previously established
- and also that
- then you can deduce
- Or if you've established
- and also that
- then you can deduce
- To prove subset ralations (where and are sets)
- Take a general member of , and give it a name ("let ")
- Use definition of to say something about
- Follow through the logical consequences of that
- aiming to prove that also satisfies the definition of
- To prove set equality "" prove and
- To prove numerical equality "" prove and
- Also known as proof by example
- Can be used where the theorem asserts the existence of some object with a specific property just give the example, show it has the property
- Also known as proof by exhaustion or (if lots of cases) "brute force"
- Identify and prove a number of different cases which cover all possibilities
- Also known as "reduction ad absurdum"
- Start by assuming the negation of the statement you want to prove
- Deduce a contradiction
- Therefore, the statement must be true
- Prove base case
- then prove inductive case where is true which implies is true
- Proof:
- Assume by way of contradiction that there are finitely many primes
- Let be the number of primes
- Let be all the primes
- Define
- This is bigger than any prime
- Therefore must be a composite of primes
- However, , if you divide by you will get a remainder of by our construction
- So cannot be a multiple of therefore cannot be a multiple of any prime number
- Thus by way of contradiction there is no largest prime
- A set is countable if either
- It is finite, or
- It can be put in one-to-one correspondence (i.e. bijection) with
- For example the set of all FIT2014 students is finite,
- The sets , and can all be put in one-to-one correspondence with
- However, the set of all sets of strings is not countable
- The sets of all languages is not countable
- Proof:
- Suppose by way of contradiction that the set of all languages is countable
- Since we know it's not finite, there must be a bijection between and {all languages}
- Let the members of the set of all languages be
- Recall, that the set of all strings is countable so we can list them as
- Define a language as follows:
- We have constructed so that, for each , it differs from in whether or not it contains
- So it differs from all languages, yet is it a language, this is a contradiction
- The set of all languages is uncountable
- Recall that : always true if is False regardless of
- |{McTaggart, The Pope}|
- |{McTaggart, The Pope}| as
- McTaggart is the Pope
- In an induction proof a lack of a base step allows for invalid proofs
- Additionally proving a base of and using allows for some strange behaviour such as with the proof that every word is uniform (made up of only one letter (lecture 2.2))
- Often long and not to the point
- However, there may be some uses like for example proving no formula using basic arithmetic operations for the 5th power polynomial or higher