P implies q and q implies p 349507-P implies q and q implies p

Solved Question 4 Suppose P Q And Rare Statements If P Is Chegg Com

Truth table (p implies q) and ((not p) implies (not q) ) Natural Language;Given "p implies q", there are two possibilities We could have "p", and therefore "q" (so q is possibility 1) Or, we could have "not p", and therefore, we would not have q (so we could use possibility 2 as not p) Thus, "p implies q" is equivalent

P implies q and q implies p

P implies q and q implies p-P then q" or "p implies q", represented "p → q" is called a conditional proposition For instance "if John is from Chicago then John is from Illinois" The proposition p is called hypothesis or antecedent, and the proposition q is the conclusion or consequent Note that p → q is true always except when p is true and q is falseNot only does this mean you actually have half as much to memorize, there's a reason for this that we'll get into when we discuss Boolean Algebras in Discrete 2!

Solved Question 4 Suppose P Q And Rare Statements If P Is Chegg Com

It implies that "P cannot be True unless Q is True" or if"Q is False, then P is False" Hence Q is a necessary condition for PP → q = (~p ∨ q) In the Principia Mathematica, the "=" denotes "is defined to mean" Using this denotation, the above expression can be read "p implies q is defined to mean that either p is false or q is true" The following truth table shows the logical equivalence of "If pQ is necessary for p;

Press p→q "if p, then q", "p implies q", "if p, q", "p only if q", "only if q, p", "p is sufficient for q", "a sufficient condition for q is p", "q if p", "q whenever p", "q when p", "q is necessary for p", "a necessary condition for p is q", "q follows from p", "q unless ¬p" The following four questions are all based on theQuestion We said that "(P implies Q) and (Q implies P)" has the same meaning as "P iff Q" Verify that with a truth table, like we did on page 34 What about "(P or Q) implies (P and Q);" does that also have the sameP only if q;

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Answer (1 of 6) (p \Rightarrow r) \lor (q \Rightarrow r) \Leftrightarrow (\neg p \lor r) \lor (\neg q \lor r) \Leftrightarrow \neg p \lor \neg q \lor r While it is tempting to interpret " P only if Q " as " P if Q;

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