1)p->q 2)not p or q 3)p 4)not p and p or q 5)contrudiction or q 6)q
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.
"abcd is not a parallelogram or it does not have any right angles." ~(P and Q) = ~P or ~Q
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
Affirmative Syllogism: All P are Q X is a P X is a Q Negative Syllogism: All P are Q X is not a Q X is not P Both syllogisms are always valid. but dont be fooled by their evil twins the fallacy of affirmation and the fallacy of negation.
P Q (/P or /Q) T T F T F T F T T F F T
Construct a truth table for ~q (p q)
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
A+
. p . . . . . q. 0 . . . . . 1. 1 . . . . . 0
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
1)p->q 2)not p or q 3)p 4)not p and p or q 5)contrudiction or q 6)q
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.
"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'
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 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