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P Q (/P or /Q) T T F T F T F T T F F T

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16y ago

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Related Questions

Construct a truth table for p and q if and only if not q?

Construct a truth table for ~q (p q)


How do you write the negation of if and then?

If p then q is represented as p -> q Negation of "if p then q" is represented as ~(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 truth table for p and negation q?

P | T T F F Q | T F T F Q' | F T F T P + Q' | F T F F The layout is the best I could do with this software. Hope it is OK.


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 can the statement "p implies q" be expressed in an equivalent form using the logical operator "or" and the negation of "p"?

The statement "p implies q" can be expressed as "not p or q" using the logical operator "or" and the negation of "p".


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


Is The inverse is the negation of the converse?

No, the inverse is not the negation of the converse. Actually, that is contrapositive you are referring to. The inverse is the negation of the conditional statement. For instance:P → Q~P → ~Q where ~ is the negation symbol of the sentence symbols.


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