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Section 1.2

Section 1.2. Propositional Equivalences. Equivalent Propositions. Have the same truth table Can be used interchangeably For example, exclusive or and the negation of biconditional are equivalent propositions:. p q p  q p  q  (p  q) T T F T F

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Section 1.2

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  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: p q p  q p  q  (p  q) T T F T F T F T F T F T T F T F F 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 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 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 p q p p  q q p  q T T F F F F T F F T T T F T T T F T F F T F T F

  9. Method 1 example Show that (p  q)  p  q p q p  q (p  q) p q p  q T T T F F F F T F F T F T T F T F T T F T F F F T T T T

  10. Method 1 example Show that p  (q  r)  (p  q)  (p  r) p q r qr p(qr) pq pr (pq)(pr) T T T T T T T T T T F T T T F T T F T T T F T T T F F F F F F F F T T T F F F F F T F T F F F F F F T T F F F F F F F 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 2N 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 -

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