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Construct a truth table for p and q if and only if not q?
Construct a truth table for ~q (p q)
Is not p and q equivalent to not p and not q?
Think of 'not' as being an inverse. Not 1 = 0. Not 0 = 1. Using boolean algebra we can look at your question. 'and' is a test. It wants to know if BOTH P and Q are the same and if they are 1 (true). If they are not the same, or they are both 0, then the result is false or 0. not P and Q is rewritten like so: (P and Q)' = X not P and not Q is rewritten like: P' and Q' = X (the apostrophe is used for not) We will construct a truth table for each and compare the output. If the output is the same, then you have found your equivalency. Otherwise, they are not equivalent. P and Q are the inputs and X is the output. P Q | X P Q | X ------ 0 0 | 1 0 0 | 1 0 1 | 1 0 1 | 0 1 0 | 1 1 0 | 0 1 1 | 0 1 1 | 0 Since the truth tables are not equal, not P and Q is not equivalent to not P and not Q. Perhaps you meant "Is NOT(P AND Q) equivalent to NOT(P) AND NOT(Q)?" NOT(P AND Q) can be thought of intuitively as "Not both P and Q." Which if you think about, you can see that it would be true if something were P but not Q, Q but not P, and neither P nor Q-- so long as they're not both true at the same time. Now, "NOT(P) AND NOT(Q)" is clearly _only_ true when BOTH P and Q are false. So there are cases where NOT(P AND Q) is true but NOT(P) AND NOT(Q) is false (an example would be True(P) and False(Q)). NOT(P AND Q) does have an equivalence however, according to De Morgan's Law. The NOT can be distributed, but in doing so we have to change the "AND" to an "OR". NOT(P AND Q) is equivalent to NOT(P) OR NOT(Q)
If P is true and Q is false what is the truth value of P or Q?
If p is true and q is false, p or q would be true. I had a hard time with this too but truth tables help. When using P V Q aka p or q, all you need is for one of the answers to be true. Since p is true P V Q would also be true:)
What is the proof for P and Not P Therefore Q?
"P and not P" is always false. If P is true, not P is false; if P is false, not P is true. In either case, combining a true and a false with the AND operator gives you false. And if you look at the truth table for the implication (the "therefore" part), when the left part is false, the result is always true.
What are the inference rules for functional dependency?
"The present list of 19 rules of inference constitutes a COMPLETE system of truth-functional logic, in the sense that it permits the construction of a formal proof of validity for ANY valid truth-functional argument." (FN1)The first nine rules of the list are rules of inference that "correspond to elementary argument forms whose validity is easily established by truth tables." (Id, page 351). The remaining ten rules are the Rules of Replacement, "which permits us to infer from any statement the result of replacing any component of that statement by any other statement logically equivalent to the component replaced." (Id, page 359).Here are the 19 Rules of Inference:1. Modus Ponens (M.P.)p qpq 2.Modus Tollens (M.T.)p q~q~p 3.Hypothetical Syllogism (H.S.)p qq rp r 4.Disjunctive Syllogism (D.S.)p v q~ pq 5. Constructive Dilemma (C.D.)(p q) . (r s)p v rq v s 6. Absorption (Abs.)p qp (p. q)7. Simplification (Simp.)p . qp 8. Conjunction (Conj.)pqp . q 9. Addition (Add.)pp v qAny of the following logically equivalent expressions can replace each other wherever they occur:10.De Morgan's Theorem (De M.) ~(p . q) (~p v ~q)~(p v q) (~p . ~q) 11. Commutation (Com.)(p v q) (q v p)(p . q) (q . p) 12. Association (Assoc.)[p v (q v r)] [(p v q) v r][p . (q . r)] [(p . q) . r] 13.Distribution (Dist) [p . (q v r)] [(p . q) v (p . r)][p v (q . r)] [(p v q) . (p v r)] 14.Double Negation (D.N.)p ~ ~p 15. Transposition (Trans.)(p q) (~q ~p) 16. Material Implication (M. Imp.)(p q) (~p v q) 17. Material Equivalence (M. Equiv.)(p q) [(p q) . (q p)](p q) [(p . q) v (~p . ~q)] 18. Exportation (Exp.)[(p . q) r] [p (q r)] 19. Tautology (Taut.) p (p v p)p (p . p)FN1: Introduction to Logic, Irving M. Copi and Carl Cohen, Prentice Hall, Eleventh Edition, 2001, page 361. The book contains the following footnote after this paragraph: "A method of proving this kind of completeness for a set of rules of inference can be found in I. M. Copi, Symbolic Logic, 5th Edition. (New York: Macmillian, 1979), chap 8, See also John A. Winnie, "The Completeness of Copi's System of Natural Deduction," Notre Dame Journal of Formal Logic 11 (July 1970), 379-382."
Construct a truth table for p and q if and only if not q?
Construct a truth table for ~q (p q)
what is the correct truth table for p V -q?
what is the correct truth table for p V~ q
what is the correct truth table for -p-> -q?
A+
Make a truth table for the statement if p then not q?
. p . . . . . q. 0 . . . . . 1. 1 . . . . . 0
How do you construct a truth table for parenthesis not p q parenthesis if and only if p?
Assuming that you mean not (p or q) if and only if P ~(PVQ)--> P so now construct a truth table, (just place it vertical since i cannot place it vertical through here.) P True True False False Q True False True False (PVQ) True True True False ~(PVQ) False False False True ~(PVQ)-->P True True True False if it's ~(P^Q) -->P then it's, P True True False False Q True False True False (P^Q) True False False False ~(P^Q) False True True True ~(P^Q)-->P True True False False
What is the truth table for p arrow q?
Not sure I can do a table here but: P True, Q True then P -> Q True P True, Q False then P -> Q False P False, Q True then P -> Q True P False, Q False then P -> Q True It is the same as not(P) OR Q
What is the proof of the modus ponens not by the truth table?
1)p->q 2)not p or q 3)p 4)not p and p or q 5)contrudiction or q 6)q
What is the truth table for negation p or negation q?
P Q (/P or /Q) T T F T F T F T T F F T
What is the truth table for p or q and the opposite of p and q?
P . . Q . . (P or Q)0 . . 0 . . . 00 . . 1 . . . 11 . . 0 . . . 11 . . 1 . . . 1=================P . . Q . . NOT(P and Q)0 . . 0 . . . . 10 . . 1 . . . . 11 . . 0 . . . . 11 . . 1 . . . . 0
P-q and q-p are logically equivalent prove?
p --> q and q --> p are not equivalent p --> q and q --> (not)p are equivalent The truth table shows this. pq p --> q q -->(not)p f f t t f t t t t f f f t t t t
Where p and q are statements p and q is called what of p and q?
The truth values.
How do you construct a truth table for q arrow p?
I guess you mean q → p (as in the logic operator: q implies p).To create this truth table, you run over all truth values for q and p (that is whether each statement is True or False) and calculate the value of the operator. You can use True/False, T/F, 1/0, √/X, etc as long as you are consistent for the symbol used for True and the symbol used for False (the first 3 suggestions given are the usual ones used).For implies:if you have a true statement, then it can only imply a true statement to be truebut a negative statement can imply either a true statement or a false one to be truegiving:. q . . p . q→p--------------. 0 . . 0 . . 1 .. 0 . . 1 . . 1 .. 1 . . 0 . . 0 .. 1 . . 1 . . 1 .
