It means the statement P implies Q.
P! / q!(p-q)!
a syllogism
This is an incomplete statement. Your question cannot be answered.
"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."
If B is between P and Q, then: P<B<Q
The statement "If not q, then not p" is logically equivalent to "If p, then q."
Converse: If p r then p q and q rContrapositive: If not p r then not (p q and q r) = If not p r then not p q or not q r Inverse: If not p q and q r then not p r = If not p q or not q r then not p r
In the statement "p implies q," the relationship between p and q is that if p is true, then q must also be true.
The statement "p if and only if q" is true when both p and q are true, or when both p and q are false.
"if p then q" is denoted as p → q. ~p denotes negation of p. So inverse of above statement is ~p → ~q, and contrapositive is ~q →~p. ˄ denotes 'and' ˅ denotes 'or'
if the statement is : if p then q converse: if q then p inverse: if not p then not q contrapositive: if not q then not
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 .
The statement "p implies q" can be expressed as "not p or q" using the logical operator "or" and the negation of "p".
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
No, it is not valid because there is no operator between P and q.
A+
P! / q!(p-q)!