Show that each of these conditional statements is a tautology by using truth tab

Question

Show that each of these conditional statements is a tautology by using truth tables. a) (p logical AND q) rightarrow p b) p rightarrow (p logical OR q) c) p rightarrow (p rightarrow q) d) (p logical AND q) rightarrow (p rightarrow q) e) (p rightarrow q) rightarrow p f) (p rightarrow q) rightarrow Show that each of these conditional statements is a tautology by using truth tables. a) [ p logical AND (p logical OR q)] rightarrow q b) [(p rightarrow q) logical AND (q rightarrow r)] rightarrow (p rightarrow r) c) [p logical AND (p rightarrow q)] rightarrow q d) [(p logical OR q) logical AND (p rightarrow r) logical AND (q rightarrow r)] rightarrow r Show that each conditional statement in Exercise 9 is as tautology without using truth tables. Show that each conditional statement in Exercise 10 is a tautology without using truth tables. Use truth tables to verify the absorption laws. a) p logical OR (p logical AND q) identical to p b) p logical AND (p logical OR q) identical to p Determine whether ( p logical AND (p rightarrow q)) rightarrow q is a tautology. Determine whether ( q logical AND (p rightarrow q)) rightarrow p is a tautology. Show that (p logical AND q) rightarrow r and (p rightarrow r) logical AND (q rightarrow r) are not logically equivalent. Show that (p rightarrow q) rightarrow (r rightarrow s) and (p rightarrow r) rightarrow (q rightarrow s) are not logically equivalent. The dual of a compound proposition that contains only the logical operators logical OR, logical AND, is the compound proposition obtained by replacing each logical OR by logical AND, each logical AND by logical OR, each T by F, and each F by T. The dual of s is denoted by s*. Find the dual of each of these compound propositions. a) p logical OR q b) p logical AND(q logical OR (r logical AND T)) c) (p logical AND q) logical OR (q logical AND F) Find the dual of each of these compound propositions . a) p logical AND q logical AND r b) (p logical AND q logical AND r) logical OR s c) (p logical OR F) logical AND (q logical OR T) When does s* = s, where s is a compound preposition? Show that (s*)* = s when s is a compound proposition. Show that the logical equivalences in Table 6, except for the double negation law, come in pairs, where each pair contains compound prepositions that are duals of each other. Why are the duals of two equivalent compound propositions also equivalent, where these compound propositions contain only the operators logical AND, logical OR and ? Find a compound proposition involving the propositional variables p, q, and r that is true when p and q are true and

Explanation / Answer

a. (~p ^ (p v q)) -> q

= ((~p ^ p) v (~p ^ q)) -> q Distributive law

= (F v (~p ^ q)) -> q

= (~p ^ q) -> q

= (q ^ ~p) -> q Commutative law

= q -> q Simplification

b. [(p->q) ^ (q->r)] -> (p->r)

=> [(~pvq) ^ (~qvr)] -> (p->r) Material implication

=> [((~pvq)^~q) v ((~pvq)^r)] -> (p->r) Distributive law

=> [((p->q)^~q) v ((~pvq)^r)] -> (p->r) Material implication

=> [~p v ((~pvq)^r)] -> (p->r)

=> [~p v r] -> (p->r) Simplification

=> (p->r) -> (p->r)

c. (p ^ (p -> q)) -> q

=> (p ^ (~p v q)) -> q Material implication

=> ((p ^ ~p) v (p ^ q)) -> q Distributive law

=> (F v (p ^ q) -> q

=> (p ^ q) -> q

=> q -> q Simplification

d. [ (p v q) ^ (p -> r) ^ (q -> r)] -> r

=> [ (~p -> q) ^ (p -> r) ^ (q -> r) ] -> r Material implication

=> [ (~p -> q) ^ (q -> r) ^ (p -> r) ] -> r Associative law

=> [ (~p -> r) ^ (p -> r) ] -> r The result of b which is known as hypothetical syllogism

=> [ (p v r) ^ (~p v r) ] -> r Material implication

=> [ (p ^ (~p v r)) v (r ^ (~p v r)) ] -> r Distributive law

=> [ (p ^ (p -> r)) v (r ^ (p -> r)) ] -> r

=> [ r v (r ^ (p -> r)) ] -> r Result of c which is known as modus ponens

=> [ (r v r) ^ (r v (p -> r)) ] -> r Distributive law

=> (r v r) -> r Simplification

=> r -> r

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