Proof strategies, or, a manual of logical style

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1 Proof strategies, or, a manual of logical style Dr Holmes September 27, 2017 This is yet another version of the manual of logical style I have been working on for many years This semester, instead of posting one of my old versions, I am going to develop the manual section by section as I lecture bits to you I m probably also going to add some coverage of things we have already talked about The basic idea behind this document is that there is a set of stereotyped strategies for proving statements of given logical forms and using statements of given logical forms which are assumed in a proof The purpose here is to list the basic ones The application is that one can form a good idea how the process of writing a proof of a given statement will start from the way the statement is written There will be a point where you have to apply some creativity to complete an argument, of course, but if you know standard strategies a lot of the work of writing a proof can be made mechanical 1 Implication Our author tells us the basic strategy for proving an implication: To prove A B assume A for the sake of argument and prove B If you succeed in proving B with the extra assumption, then you have proved A B An outline is given on the next page 1

2 Goal: A B Assume (1): A Goal: B (these dots stand for many proof steps) (n): B (n+1): A B, by deduction, lines 1 n It is important to notice that the indented lines from the assumption of A to the conclusion B will not be referred to again in the proof, since they depend on the assumption A made strictly for the sake of argument, and so are no longer available once we are no longer making that assumption Notice that we give this proof strategy the name deduction There will be another proof strategy for proving implications which we will discuss later To use an assumption which is an implication the basic rule has the Latin name modus ponens: A A B B If we can assume A and we can assume A B, then we can conclude B This is an example of a convenient general format for a logical rule: if one has statements of the form listed above the line, one can deduce the statement written below the line There is another rule for using an implication, modus tollens, which we will see in due course 2

3 2 Conjunction The rules for conjunction are so transparent that you may not realize you are doing anything when you use them To prove a conjunction A B, prove A then prove B This can be called the rule of conjunction If you can assume A B, you can further conclude A If you can assume A B, you can further conclude B The last two rules can be called rules of simplification In the handy format given above, conjunction is and the rules of simplification are A B A B and A B A A B A 3 Biconditional A statement A B is equivalent to (A B) (B A), and this equivalence indicates how to prove it: first prove A B then prove B A Here is an outline (next page): 3

4 Goal: A B Part I: Assume (1): A Goal: B (n): B (The dots stand for many proof lines) Part II: Assume (n+1): B Goal: A (many proof steps) (n+k): A (n+k+1): A B, biconditional proof, 1-n, n+1-n+k We also have these rules to use a biconditional, suspiciously like modus ponens: P P Q Q Q P Q P 4 A sample proof using just implication, conjunction and biconditional You may recall that we proved this theorem using truth tables Main Goal: Prove ((P Q) R) (P (Q R)) This statement is a biconditional, so its proof will have two parts Part I begins on the next page 4

5 Part I: Assume (1): ((P Q) R) Goal (of part I): (P (Q R)) The form of the goal tells us what to do! Assume (2): P Goal: Q R and again! Assume (3): Q Goal: R Now we have unpacked everythingwe should pause and take stock of what we have Line (1), ((P Q) R), would give us our goal Rif we had P Q And we can have this (4): P Q by conjunction, lines 2,3 (5): R by modus ponens, lines 1,4 [I do not care which order the line numbers are given in] (6): Q R by deduction lines 3-5 [this is by the entire block of lines, I do mean a hyphen not a comma] (7): (P (Q R)) by deduction lines 2-6 This completes our work for Part I We could state an implication as proved here, but we do not have to Part II is on the next page 5

6 Part II: Assume (8): (P (Q R)) note that while I numbered this line 8, to avoid conflict with the part of the proof already given, I in fact have no lines available to me but line 8 for my argument at this point: the argument of Part I is over, and everything in lines 1-7 depends on the assumption in line 1 which we are no longer making here Goal (of part II): ((P Q) R) The form of the goal tells us what to do! Assume (9): P Q Goal: R We are now as unpacked as we can get, so we have to look at our resources If we had P, we could apply modus ponens with line 8 and get Q R And we can have P (10): P simplification line 9 (11): Q R mp lines 9,10 (mp is an allowed abbreviation for modus ponens ) Since we have Q R, it is natural to think about whether we have Q And we do (12): Q simplification line 9 [we could have unpacked this at the same time we unpacked P above: the line order here is a little flexible] (13): R by mp lines 11,12 (14): ((P Q) R) by deduction lines 9-13 This completes the argument for Part II (15): The Main Goal to be proved follows by biconditional introduction, lines 1-7, 8-14 I could say ((P Q) R) (P (Q R)) follows by biconditional introduction, lines 1-7, 8-14, but this is a situation where it is natural (since the statement of the Main Goal is long) to refer to it by name 6

7 5 Negation To make presentation of the rules for negation easier, we introduce a special symbol, which we read the absurd, for a false statement First of all, we have the rule of double negation: P P We can argue from P to P as well, but this is not a basic rule (we can prove it) We have the rule of contradiction: P P If we have assumed both P and P we are on the wrong track! We have the proof technique which we call negation introduction : to prove P, assume P and deduce Goal: P Assume (1): P Goal: (you can write contradiction where your goal is to prove an absurdity) (n): (many proof lines) (n+1): P by negation introduction, lines 1-n 7

8 This is not quite the same as the famous proof technique proof by contradiction or reductio ad absurdum: Goal: P Assume (1): P Goal: (you can write contradiction where your goal is to prove an absurdity) (n): (many proof lines) We could say here that we have proved P by negation introduction then P by double negation, but we shorten this to (n+1): P by reductio ad absurdum, lines 1-n The strategy of proof by contradiction is unusual because it doesn t rely on the statement to be proved having any particular logical form It is often effective as a last resort when you see nothing else to do Notice that it is not a basic rule: it is derivable from negation introduction and double negation But it s very useful From a contradiction, anything follows We already know that this is our official position, from the truth table for implication Here is how we prove it here Main Goal: P Assume (1): Goal: P Assume (2): P (for proof by contradiction) Goal: (3): copied from line 1 (4): P reductio ad absurdum 1-3 (5): Main goal, deduction, lines 1-4 8

9 We give the (too?) easy proof that given P we can deduce P Assume (1): P Goal: P We set up for a proof by negation introduction: Assume (2): P Goal: (3):, by contradiction, 1,2 (4): P negation introduction lines 2-3 You may use this result and call it double negation as well (if you have P as a line, you may introduce P as a line, and vice versa) 9

10 6 Disjunction (or) The basic rules for proving a disjunction are only useful in special circumstances We can prove P Q (obviously) if we can either prove P or prove Q P P Q Q P Q These two rules are called the rules of addition The basic rule for using an assumption which is a disjunction is proof by cases Suppose that we have assumed P Q, and from P we can prove R and from Q we can prove R: since in either case R follows, we can tell that R follows from P Q (without needing to decide which of P or Q is true) The format is presented on the next page 10

11 The format for a proof by cases is as follows: (1): P Q Goal: R Case 1: assume (k): P Goal: R (many proof steps) (k+n): R at this point we will not refer to individual statements in the block k-k+n again, as they rely on the special assumption P Case 2: assume (k+n+1): Q Goal: R (many proof steps) (k+n+m+1): R (k+n+m+2): R, proof by cases, lines 1, k-k+n, (k+n+1)-(k+n+m+1) This rule has the most complex line justification of any of the rules of the system (a line and two blocks of lines) The weird line numbering is due to the fact that of course we cannot tell how long the blocks will be Also, I want to signal clearly that the disjunction (line 1) can be used for proof by cases much later in the proof (so the next line we see is line k rather than line 2) 11

12 7 Some proofs of useful theorems We prove the equivalence of an implication with its contrapositive This proof justifies the use of more powerful rules combining implication and negation which are given in the next section Theorem: (P Q) ( Q P ) Observations: We need to prove the implication (P Q) ( Q P ) (Part I) then the implication ( Q P ) (P Q) (Part II) Part I: Assume (1): P Q Goal: Q P Assume (2): Q Goal: P (We set up for a proof by negation introduction) Assume (3): P Goal: (4): Q by 1,3 modus ponense (5): 2,4 contradiction (6): P 3-5 negation introduction (7): Q P 2-6 deduction (8): (P Q) ( Q P ) deduction 1-7 We continue with the proof of Part II on the next page 12

13 Part II: Assume (9): Q P Goal: P Q Assume (10): P Goal: Q (At this point we may feel that we have no leverage on Q at all: this is a good point to try reduction ad absurdum) Assume (11): Q Goal: (12): P 9,11 mp (13): 10, 12 contradiction (14): Q reductio ad absurdum (15): P Q deduction (16): ( Q P ) (P Q) deduction 9-14 (17): The main theorem (P Q) ( Q P ), by biconditional introduction 8, 15 (this is an alternative style of justifying an if and only if proof, using explicit references to the two implications instead of references to the blocks which prove them: the usual style would be biconditional introduction 1-7, 9-15; you may use either approach) 13

14 8 Additional rules combining implication and disjunction with negation Additional rules for implication follow from the equivalence of P Q with Q P This equivalence can be verified with a truth table it can also be proved with the earlier given rules, as we showed in the previous section, and this proof can be used to justify the new rules combining implication and negation that we give Additional rules for disjunction follow from the equivalence of P Q with both P Q and Q P (you can check this with a truth table; these equivalences are also provable using the earlier given rules) The rule of proof by contrapositive for implication: Main Goal: P Q We set up to prove Q P instead: Assume (1): Q Goal: P (many proof steps) (n): P (n+1): P Q, proof by contrapositive 1-n The rule of modus tollens: P Q Q P You should notice that both of the preceding rules are derived simply by reading P Q as equivalent to Q P 14

15 We present our usual format for proving a disjunction: Main Goal: P Q The key is that we are setting up a proof of P Q Assume (1): P Goal: Q (many proof steps) (n): Q (n+1): P Q alternative exclusion, lines 1-n The name alternative exclusion is mine: we exclude an alternative and show that the other follows An alternative approach (this has nothing to do with cases, you do not need to write both of these proofs to prove P Q, just one of them) Main Goal: P Q The key is that we are setting up a proof of Q P Assume (1): Q Goal: P (many proof steps) (n): P (n+1): P Q alternative exclusion, lines 1-n Here is a famous theorem (excluded middle) with a blindingly short proof: Goal: P P Assume (1): P Goal: P (2): P P alternative exclusion 1-1 One could copy line 1 to a line 2 and have alternative exclusion 1-2, but it seems to me that when the assumption is the goal one can stop right there! 15

16 Additional rules for use of disjunctions, called rules of disjunctive syllogism are derived from the same equivalences: P Q P Q P Q Q P P Q P Q P Q Q P The first two are the actual rules of disjunctive syllogism: the second two may also be called disjunctive syllogism, but also involve an application of double negation They happen often in formal proofs, so I allow them to be viewed as single steps 16

17 9 A proof of a disjunction from algebra Zero factor theorem: If xy = 0 then x = 0 or y = 0 Proof: Let x, y be arbitrary Assume x 0 Then xy = 0, so x 1 (xy) = x 1 0, so (x 1 x)y = 0, so 1y = 0, so y = 0 Thus x = 0 or y = 0 The reason that this example is here is that you should notice this logical form: Goal: x = 0 y = 0 Assume(1): x = 0 (some proof steps) (n): y = 0 (n+1): x = 0 y = 0 (by alternative exclusion) And more generally, the point of all these forms of argument is that they really happen, in English, in mathematical arguments in this book and in all subsequent books you will read This is not to say that my particular presentation of the rules is universal: there are lots of different ways to articulate the rules of logic 17

18 10 Quantifiers 18

19 11 Exercises 1 Prove ((P Q) (Q R)) (P R) using the style given here This should be straightforward 2 Prove ((P R) (Q R)) (Q P ) You will want to use the additional rules involving negation 3 Prove ((P R) (Q R)) ((P Q) R) This should be relatively straightforward proof by cases is needed 4 Prove ((P Q) ( Q R)) (P R) I see one approach using excluded middle and proof by cases, but I think there is a simpler way using the rules of alternative exclusion and disjunctive syllogism 5 Prove ((P Q) R) ((P R) (Q R)) This is quite tricky You should recognize this problem and a previous one as the two directions of a biconditional you proved using truth tables 6 Prove (P Q) ( P Q) This is one of demorgan s laws Notice that this is a biconditional so you have two directions of argument to complete You can t use a demorgan law to prove it; just the rules in this handout 19