1 Section 1.2 Propositional Equivalences

2 Equivalent Propositions Have the same truth table Can be used interchangeably For example, exclusive or and the negation of biconditional are equivalent propositions: pq p  qp  q  (p  q) TT F T F TF T F T FT T F T FF F T F

3 Equivalent propositions Logical equivalence is denoted with the symbol  If p  q is true, then p  q

4 Tautology A compound proposition that is always true, regardless of the truth values that appear in it For example, p  p is a tautology: p pp p  p T F T F T T

5 Contradiction A compound proposition that is always false For example, p   p is a contradiction: p pp p   p T F F F T F

6 Tautology vs. Contradiction The negation of a tautology is a contradiction, and the negation of a contradiction is a tautology Contingency: a compound proposition that is neither a tautology nor a contradiction

7 Determining Logical Equivalence Method 1: use truth table Method 2: use proof by substitution - requires knowledge of logical equivalencies of portions of compound propositions

8 Method 1 example Show that  p  q  p   q pq  p  q  q p  qq TT F F F F TF F T T T FT T T F T FF T F T F

9 Method 1 example Show that  (p  q)   p   q pqp  q  (p  q) pp qq  p  qq TT T F F F F TF F T F T T FT F T T F T FF F T T T T

10 Method 1 example Show that p  (q  r)  (p  q)  (p  r) pqr qrqrp  (q  r)pqpqprpr(p  q)  (p  r) TTT T T T T T TTF T T T F T TFT T T F T T TFF F F F F F FTT T F F F F FTF T F F F F FFT T F F F F FFF F F F F F

11 The limits of truth tables The previous slide illustrates how truth tables become cumbersome when several propositions are involved For a compound proposition containing N propositions, the truth table would require 2 N rows

12 Method 2: using equivalences There are many proven equivalences that can be used to prove further equivalences Some of the most important and useful of these are found in Tables 5, 6 and 7 on page 24 of your text, as well as on the next several slides

13 Identity Laws p  T  p p  F  p In other words, if p is ANDed with another proposition known to be true, or ORed with another proposition known to be false, the truth value of the compound proposition will be the truth value of p

14 Domination Laws p  T  T p  F  F A compound proposition will always be true if it is composed of any proposition p ORed with any proposition known to be true. Conversely, a compound proposition will always be false if it is composed of any proposition p ANDed with a proposition known to be false

15 Idempotent Laws p  p  p p  p  p A compound proposition composed of any proposition p combined with itself via conjunction or disjunction will have the truth value of p

16 Double negation  (  p)  p The negation of a negation is … well, not a negation

17 Commutative Laws p  q  q  p p  q  q  p Ordering doesn’t matter in conjunction and disjunction (just like addition and multiplication)

18 Associative Laws (p  q)  r  p  (q  r) (p  q)  r  p  (q  r) Grouping doesn’t affect outcome when the same operation is involved - this is true for compound propositions composed of 3, 4, 1000 or N propositions

19 Distributive Laws p  (q  r)  (p  q)  (p  r) p  (q  r)  (p  q)  (p  r) OR distributes across AND; AND distributes across OR

20 DeMorgan’s Laws  (p  q)   p   q  (p  q)   p   q The NOT of p AND q is NOT p OR NOT q; the NOT of p OR q is NOT p AND NOT q Like Association, DeMorgan’s Laws apply to N propositions in a compound proposition

21 Two Laws with No Name p   p  T p   p  F A proposition ORed with its negation is always true; a proposition ANDed with its negation is always false

22 A Very Useful (but nameless) Law (p  q)  (  p  q) The implication “if p, then q” is logically equivalent to NOT p ORed with q

23 Method 2: Proof by Substitution Uses known laws of equivalences to prove new equivalences A compound proposition is gradually transformed, through substitution of known equivalences, into a proveable form

24 Example 1: Show that (p  q)  p is a tautology 1. Since (p  q)  (  p  q), change compound proposition to:  (p  q)  p 2. Applying DeMorgan’s first law, which states:  (p  q)   p   q, change compound proposition to:  p   q  p 3. Applying commutative law: p   p   q 4. Since p   p  T, we have T   q 5. And finally, by Domination, any proposition ORed with true must be true - so the compound proposition is a tautology

25 Example 2: Show that  p  q and p   q are logically equivalent 1. Start with definition of biconditional: p  q  p  q  q  p; then the 2 expressions become: (  p  q)  (q   p) and (p   q)  (  q  p) 2. Since p  q   p  q, change expressions to: (  (  p)  q)  (  q   p) and (  p   q)  (  (  q)  p); same as:(p  q)  (  q   p) and (  p   q)  (q  p) 3. Reordering terms, by commutation, we get: (p  q)  (  p   q) and (p  q)  (  p   q) Since the two expressions are now identical, they are clearly equivalent.

26 Section 1.2 Propositional Equivalences - ends -