Propositional Logic

Argument Forms

( )

( )

( )

( )

If we further define an equivalence connective for any two propositions p and q by (p q) (q p), denoted by p q, then the usual ''order of precedence'' is

1.

connectives within parentheses, innermost parentheses first

2.



3.

,

4.



5.

Example

1. p q r p is same as p ((q ( r)) p). However p q r p would be same as p (q (( r) p)) had we adopted higher precedence for than for .

An argument form, or argument for short, is a sequence of statements. All statements but the last one are called premises or hypotheses. The final statement is called the conclusion, and is often preceded by a symbol '' ''.

An argument is valid if the conclusion is true whenever all the premises are true.

Examples

2. Show (p q, p r, q r, r) is a valid argument.

   Solution From the table

   

   we see all critical rows (in this case, those with the shaded positions all containing a T) correspond to (the circled) T(true) for r. Hence the argument is valid.

3. Show that the argument (p q, p q ) is invalid.

   Solution

   

   We see that on the 3rd row, a critical row, the premise p q is true while the conclusion p q is false. Hence the argument (p q, p q ) is invalid.

Converse, Inverse and Contrapositive

p q

q p is the converse of p q;

p q is the inverse of p q;

q p is the contrapositive of p q.

4. Let p be the statement ''I'm sick'' and qbe the statement ''I go and see a doctor''. The p qmeans ''If I'm sick, then I go and see a doctor''. The converse of this statement is ''If I go and see a doctor, them I'm sick'', while the inverse reads ''If I'm not sick, then I won't go and see a doctor''. However, the contrapositive takes the form ''If I don't go and see a doctor, then I'm not sick''.

	converse error:	p q,   q p			invalid argument forms
	inverse error:	p q,   p q

Rules of Inference

modus ponens:      p q, p,   q

modus tollens:      p q, q,   p

shows the validity of the argument form.

disjunctive addition:      p,   p q.

conjunctive addition:      p, q,   p q.

conjunctive simplification:      p q,   p

disjunctive syllogism:      p q, q,   p

hypothetical syllogism:      p q, q r,   p r

division into cases:      p q, p r, q r,   r

rule of contradiction:      p contradiction,   p

Examples

5. Show that the following argument form

   p q, q r, p s t, r, q u s,   t

   (*)

   is valid by breaking it into a list of known elementary valid argument forms or rules.

   Solution We'll treat all the rules of inference introduced earlier in this subsection as the known elementary argument forms. The logical inference for the argument form in the question is as follows.

   (1)	q r	premise
   	r	premise
   	q	by modus tollens
   (2)	p q	premise
   	q	by (1)
   	p	by disjunctive syllogism
   (3)	q u s	premise
   	q	by (1)
   	u s	by modus ponens
   (4)	u s	by (3)
   	s	by conjunctive simplification
   (5)	p	by (2)
   	s	by (4)
   	p s	by conjunctive addition
   (6)	p s t	premise
   	p s	by (5)
   	t	by modus ponens

6. Can you explain, in additional details, how those 6 proof steps in the above example come into existence?

   Solution In the following we shall give the ''reverse route'' for the proof of the above argument form. Basically we shall start with the conclusion t (marked by below), then go downwards towards other ''required'' intermediate propositions by making use of the known premises. The final proof of the argument form will be essentially in the reverse order of the ''reverse route''. The main idea is to try to lead from the conclusion to eventually reach only the premises.

   

   From the above diagram we see that we could derive the conclusion tin the order of

   

   With the help of the above order of derivation, we have

   (a)	r, q r,   q	(modus tollens)
   (b)	q, p q,   p	(disjunctive syllogism)
   (c)	q, q u s,   u s	(modus ponens)
   (d)	u s,   s	(conjunctive simplification)
   (e)	p, s,   p s	(conjunctive addition)
   (f)	p s, p s t,   t	(modus ponens)

   That is, the conclusion is derived from the use of the basic inference rules.

7. Can you do example 6 once agagin, with some differences?

   Solution The ''dumbest'' way is to try to determine for each proposition (symbol) if it is true or not. This way one could be wasting a lot of time unnecessarily; but this often ensures one gets closer and closer to the solution of the problem.

   In the argument form in examples 5-6, we have 6 basic propositions p, q, r, s , t and u. We will now proceed to determine (in the order dictated by the actual circumstances) whether each of these 6 propositions is true or not.

   (i)

   r is given as a premise, so the truth value of proposition r is already determined (r is false).

   (ii)

   q r, r,   q,    so the truth value of q is determined (q is false).

   (iii)

   p q, q,   p,    so the truth value of p is determined (p is true).

   (iv)

   q, q u s,   u s.

   (v)

   u s,   u,    so u is determined (u is true).

   (vi)

   u s,   s,    so s is determined (s is true).

   (vii)

   p, s,   p s.

   (viii)

   p s, p s t,   t,    so the truth value of t is determined (t is true).

   We have by now established the truth value of all the concerned basic propositions p,q,r,s,t,u. We can now proceed to determine if the conclusion t of the original argument form is true or not. In this case, it is obvious because we have already shown t is true. So the argument form (*) in example 5 is valid. Notice that step (v) is completely unnecessary. But we probably wouldn't know it at that time.

   Note This ''dumb'' method can be very useful if you want to determine whether or not the conclusion of a lengthy argument form is true or not. I'll leave you to think about why.

8. A detective established that one person in a gang comprised of 4 members A,B,C and D killed a person named E. The detective obtained the following statements from the gang members (S_A denotes the statement made by A. Likewise for S_B, S_Cand S_D)

   (i)

   S_A:    B killed E.

   (ii)

   S_B:    C was shooting craps with A when E was knocked off.

   (iii)

   S_C:    B didn't kill E.

   (iv)

   S_D:    C didn't kill E.

   The detective was then able to conclude that all but one were lying. Can you decide who killed E?

   Solution Let (1)--(4) to be given later on be 4 statements. The 1st two, (1) and (2) below, are true due to the detective's work.

   (1):

   Only one of the statements S_A, S_B, S_C, S_D. is true

   (2):

   One of A, B, C and D killed E.

   From the (content of the) statements S_A and S_C we know

   (3):

   S_A S_D is true because if S_A is true, then B killed E which implies C didn't kill E due to (2), implying S_D is also true.

   and from (1)

   (4):

   S_A S_B S_C S_D is true.

   Let us examine the following sequence of statements.

   (a)	SA SD	SA true implies SDtrue, i.e. (SA, SD) (from (3))
   (b)	SA SB SC SD	SA true implies SB, SC and SDall false (from (4))
   (c)	SB SC SD SD	conjunctive simplification (direct)
   (d)	SA SD	hypothetical syllogism (from (b), (c))
   (e)	SD SA	modus tollens (from (a))
   (f)	SA SA	hypothetical syllogism (from (d), (e))

   We note all the statements on the sequence apart from the first two (a) and (b) are obtained from their previous statements or form the valid argument forms. However the first 2 statements (a) and (b) are both true hence the conclusion in (f) is also true. A statement sequence of this type is sometimes called a proof sequence with the last entry called a theorem. The whole sequence is called the proof of the theorem.

   Alternatively sequence (a)--(f) can also be regarded as a valid argument form in which a special feature is that the truth of the first 2 statements will ensure that all the premises there are true.

   From (3) and (4) and (a)--(f) we conclude S_A S_A is true. Hence S_A must be false from the rule of contradiction (if S_A were true then S_A would be true, implying S_A is false: contradiction).

   From the definition of S_C we see S_A S_C. From the modus ponens

   S_A S_C, S_A,   S_C

   we conclude S_C is true. Since (1) gives

   S_C S_A S_B S_D ,

   we obtain from the (conjunctive simplification) argument

   S_C S_A S_B S_D, S_A S_B S_D S_D,   S_C S_D

   that S_C S_D. Finally from modus ponens (S_C S_D, S_C,   S_D) , we conclude S_D is true, that is, C killed E. We note that in the above example, we have deliberately disintegreted our argument into smaller pieces with mathematical symbolisation. It turns out that verbal arguments in this case are much more concise. For a good comparison, we give below an alternative solution.

9. Re-do the previous qestion more directly.

   Solution (alternative for example 6) Suppose A wasn't lying, then A's statement B killed E is true. Since A spoke the truth means B,C and D would be lying, hence the statement C didn't kill E said by D would be false, implying C did kill E. But this is a contradiction to the assumption A spoke the truth. Hence A was lying, which means B didn't kill E, which in turn implies C spoke the truth. Since only one person was not lying, D must have lied. Hence Cdidn't kill E is false. Hence C killed E.