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# T1.2 - Will be done in class 5 Proof of Logical Equivalence...

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TAUTOLOGY: A compound proposition that is TRUE for all combinations of its atomic propositions E.g. (p/\q) -> (p\/q) p q p/\q p\/q (p/\q) -> (p\/q) T T T T T T F F T T F T F T T F F F F T CONTRADICTION: A compound proposition that is FALSE for all combinations of its atomic propositions E.g. –p /\ (p/\q) p q p/\q -p -p/\(p/\q) T T T F F T F F F F F T F T F F F F T F 1

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LOGICAL EQUIVALENCE (p<=>q): Two compound propositions p and q are logically equivalent iff their biconditional p <-> q is a tautology. E.g. -(p \/ q) <=> -p/\-q (De Morgan’s law) p q -p -q p\/q -(p\/q) -p/\-q -(p \/ q)<-> -p/\-q T T F F T F F T T F F T T F F T F T T F T F F T F F T T F T T T 2
Prove the distributive law using “compact” truth tables: p \/ (q /\ r) <=> (p \/ q) /\ (p \/ r) p \/ (q /\ r) <-> (p\/q) /\ p\/r T T T T T T T T T T T T F F T T T T T T F F T T T T T T T F F F T T T T F T T T T T T T T F F T F F T T F F F F F F T T F F T F F F F F T F F F 3

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A result we will use later is: p <-> q <-> ( ( p -> q ) /\ ( q -> p ) ) T T T T T T T T T T T T F F T T F F F F T T F F T T F T T F T F F F T F T F T F T F T F This proves that p<->q<=>( (p->q)/\(q->p) ) 4
[[In Class Exercise]] Prove that p->q <=> -p\/q

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Unformatted text preview: Will be done in class. 5 Proof of Logical Equivalence by Derivation (These use the Table of Logical Equivalences (6,7,8 [5,6,7 in ed.5]) in Section 1.2 plus other known Equivalences as "reasons" for steps in the proof. See Examples 6 and 7 (5 and 6 in ed. 5) for further illustrations) Prove: p -> p\/q is a tautology: 1. p-> p\/q Starting point 2. -p \/ (p\/q) Implication law (Tbl 6[5 ed. 5]) 3. (-p\/p) \/ q Associative laws 4. T \/ q-p\/p is a known tautology 5. T Identity laws QED Prove -p<->q <=> p<->-q: 1. -p<->q Starting point 2. (-p->q) /\ (q->-p) Proven in class 3. (-q->--p) /\ (--p->-q) Contrapositive Law 4. (-q->p) /\ (p->-q) Double Negation Law 5. (p->-q) /\ (-q->p) Commutative Laws 6. p<->-q Same reason as in 2. 6...
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T1.2 - Will be done in class 5 Proof of Logical Equivalence...

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