1. Activity 1-19: The Propositional Calculus www.carom-maths.co.uk

2. Consider the following Truth Table: If we take P and Q as statements, and then read the ‘upside down v’ as ‘and’, we have the proposition P and Q is true if and only if P is true and Q is true. This makes sense when compared to our daily usage of ‘and’. Task: draw up possible Truth Tables for P or Q, P implies Q, P if and only if Q, and discuss these. Here is the table for ~, which could stand for ‘not’.

3. P or Q is straightforward. P if and only if Q is also clear. P implies Q, however, is a little tricky - in fact, here we pick the following Truth Table; This looks counter-intuitive. The top half of the table is okay, but how can something false imply something true? Maybe it matters not that this sign  does not correspond exactly with our everyday English usage.

4. We can begin to string these propositions together using connectives like this: First, we fill in the possible truth values for P and Q (1, 3, 6, and 9). Then we fill in the truth values for the connectives 2, 5 and 8, followed by the truth values for 7 (looking at columns 5 and 8) and finally the truth values for 4 (looking at the values in 2 and 7.) You notice that here the final calculation (4) gives a truth value of ‘true’, whatever the truth values for P and Q. Something that is always true like this we call a ‘tautology’.

5. Task: work out the Truth Table for ((P  Q)^(Q  P))  (P  Q) This too is a tautology, which gives us more confidence in our choice of Truth Table for the symbol . Of course, not all Truth Tables represent tautologies: for example -

6. So let’s recap on what we have so far. 1. We have variables (that could represent statements) P, Q, R etc. that are Type-1 or Type-2 (True or False). 2. We have some connectives to go with these. From now on we will restrict ourselves to ~ (which could stand for NOT), v (which could stand for OR) ^ (which could stand for AND) and  (which could stand for IMPLIES). Notice that the idea here is to become more abstract; the everyday meanings of these signs are heading into the background.

7. 3. We can also add some rules for forming legitimate formulas, using the bracket signs, ‘(‘ and ‘)’. ((P^Q)  R)  (P  Q) obeys the grammar, ((^PQ)  (R)  (PQ  ) does not. We can say that if P and Q are legitimate formulas, then ~P, P ^ Q, P v Q and P  Q are all legitimate formulas too. (We may need to add some brackets.)

8. 4. We can now add two Transformation Rules (rules of inference). a)The Rule of Substitution: given a legitimate formula containing P, we can replace every occurrence of the formula P with any other legitimate formula, and our formula remains legitimate. So in ((P^Q)  R)  (P  Q) we can replace P with (R  Q) to get (((R  Q) ^Q)  R)  ((R  Q)  Q).

9. We are slowly building a logical structure here, to which we can now add some axioms. Our second Rule of Transformation is b) The Rule of Detachment: given P and P  Q, then we also have Q. 5. Our axioms are: a)(P v P)  P b)P  (P v Q) c)(P v Q)  (Q v P) d)(P  Q)  ((R v P)  (R v Q)) It is easy to see how we could use our Rules of Transformation with our axioms to infer further legitimate formulae – these are then called ‘theorems’.

10. Now let’s consider whether or not this logical system is CONSISTENT; is it possible to deduce two contradictory theorems P and ~P starting from these axioms? How to start on this? Well, it is a theorem that P  (~P  Q). Task: can you show that this is a theorem using the rules you have been given? What are the consequences of this? If P and ~P are both theorems, then Q is a theorem, and so any legitimate formula in the system is a theorem.

11. Our strategy from here goes like this: 1. find a property that is common to all four axioms. So if the system is consistent, every legitimate formula is a theorem; or putting it the other way, if we can find a legitimate formula that is not a theorem, then the system is consistent. 2. Show that this property is hereditary under our Transformation Rules, so every theorem has it. 3. Find a legitimate formula that does NOT have the property.

12. What could this property be? Well, we can return to the idea of a TAUTOLOGY. It is easy to check that all the axioms are tautologies, Using our Truth Table method. It can be shown that the Transformation Rules preserve the property of being a tautology. We know that there are legitimate formulae that are not tautologies; we showed this earlier for ((P  Q)^Q)  P. And so we have proved that our system is consistent.

13. It can be shown the answer is ‘Yes’. So here we have a simple logical system where we can prove relatively straightforwardly both completeness and consistency. What happens if we make our system more complicated? Suppose we try to add enough in the way of axioms and rules so that our system includes arithmetic – can we still prove consistency and completeness then? You might ask if our system is also complete; that is, can every possible tautology be proved from it? In 1930, the mathematician Godel proved an amazing result...

14. Kurt Godel, Austrian-American, (1906-78) Any system complicated enough to include arithmetic cannot be both consistent and complete. Or putting this another way; in any consistent system of mathematics complicated enough to include arithmetic, there will always be true statements that cannot be proved. This was a shock at the time (and still is...)

15. With thanks to: Nagle and Newman for their book Godel’s Proof. Carom is written by Jonny Griffiths, hello@jonny-griffiths.nethello@jonny-griffiths.net