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+36 votes
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Consider the following expressions:

  1. $false$
  2. $Q$
  3. $true$
  4. $P\vee Q$
  5. $\neg Q\vee P$

The number of expressions given above that are logically implied by $P \wedge (P \Rightarrow Q)$ is ___________.

 

asked in Mathematical Logic by Veteran (49.5k points)
edited by | 4.4k views
how v?
$(P \;\wedge (P\Rightarrow Q))\Rightarrow ( Q' \vee P)$ resulting in tautology. I am not sure this way is correct.
4 statements are correct because logic implication means my conclusion follows from my arguments,so arguments--> conclusion is a tautology. so except i) all are tautologies,if u simplify it.

5 Answers

+89 votes
Best answer

$4$ should be the correct answer.

 

P Q $P\implies Q$ $P\wedge(P\implies Q)$
F F T F
F T T F
T F F F
T T T T

Suppose $\left(P\wedge(P\implies Q)\right)\iff A$ (For notational convenience)

Thus for options, $(i),(ii),(iii),(iv),(v)$
If $(A\implies \text{option x})$ is a tautology.
then $P\wedge(P\implies Q)$ logically implies $\text{option x}$
else $P\wedge(P\implies Q)$ does not logically implies $\text{option x.}$

P Q A Option(i) Option(ii) Option(iii) Option(iv) Option(v)
      False $A\Rightarrow F$ Q $A\Rightarrow Q$ True $A\Rightarrow True$ $(P\vee Q)$ $A\Rightarrow \!\!(P\vee Q)$ $\neg Q\vee P$

$A\Rightarrow$$ 
(\neg Q\vee P)$

F F F   T   T   T   T   T
F T F   T   T   T   T   T
T F F   T   T   T   T   T
T T T F F T T T T T T T T

Answer=4

P.S: Blank entries in the above truth table are like don't care conditions because in those rows the value of $A$
is set to False. Hence $(A\implies \text{Anything})$ would be set to True.

 

answered by Veteran (14k points)
edited by
Grangrats Anurag for giving best Solution.
thanks bro..
Excellent!
Actually I got confused in question..I need to improve my English!
(implied by ... confused me)
excellent!!
can anyone plzz tell me " logically implied" means...??
Great explanation Thank you so much sir
Nice explanation using truth tables,but in gate like exam we should follow Validity checking method or Tautology method.
I did not get question what is asked nd what toanswer will u pls explain
+26 votes
Without making truth table we can solve very easily

P˄(P->Q)

=P˄(~P˅Q)

=(P˄~P)˅(P˄Q)

=P˄Q

Now Check these implications whether they are right or wrong.For checking try T -> F combination if any of them generate this combination then it will be ruled out. If you see carefully  1.P˄Q -> FALSE will ruled out

1.P˄Q->FALSE

2.P˄Q->Q

3.P˄Q->TRUE

4.P˄Q->P˅Q

5.P˄Q->~Q˅P

so answer is 4
answered by Boss (7k points)
edited by

@vnc I also thought the same way of solving given expression without truth table and also got :  P ^ Q

but after that what to do ? how and what  to check???? I did not get the question

@shweta after calculation of p^Q you calculate (p^Q) -> false is a tautology or You can also search that if LHS(P^Q) is True then R will not be false because that is the only condition when implied is false.
+10 votes
ans :4

all except the option i satisfies.

P and Q -> true, P or Q,~Q or P,Q
answered by Active (2.3k points)
As per modus ponens, when Q is true, how come ~Q will be true ?

Answer is 3.

As per modus ponens Q will be true.

Since P and Q both are true, as per addition rule P V Q and ~Q V P will be true.
are u sure??? I did it as false
$p \wedge q \implies p \vee \neg q$ is a tautology.

I am little doubtful about "true". Even if "anything -> true" is always true, "true" is a literal / constant value in logic, can we "imply" it? A proposition can be true or false, but true / false, taken alone, are not propositions.

Q,  (P V Q) and (~Q V P) are propositions. Are true and false propositions? If yes, then ok. If not, then how can we logically imply some thing which is not even a proposition !

are you sure sir..I did the same
yes. 4 is the answer - unless I miss something..
The point is p^(p->q) is implied q by using modus ponens but not equivalent to q.

p^(p->q) equivalent to (p^q)  so by simplification rules both p,q are true.Hence (~qvp ) is also true.
+9 votes

  4 is the answer.

 

answered by (395 points)
edited by
+5 votes
GIVEN

P->Q

P

________

1.implies Q

2 NOW Q implies P OR Q

3.P IS TRUE so it implies P OR ~Q

4 THIS IS DIFFERENT STILL EASY....P->Q is true ..Q is true and TRUE IS TRUE (trivially)..to understand ..make table of p->T

Answer => 4
answered by Active (2k points)
edited by
bhai answer bhi bata dete ek bar :) , 0? 1? 2? 3? 4? 5?
4 -answer
Answer:

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