Disproving the Assertion: P |= (Q v R) ≠ (P |= Q) or (P |= R)

[EvLer]

I somehow need to prove that a and b are not necessarily the same:
a. P |= (Q v R)
b. either (P |= Q) or (P |= R) or both

I am not asking for a full solution, just directing hints...
thanks in advance

 

[HallsofIvy]

In less technical language, (a) says "P implies Q or R" or "If P is true then either Q is true or R is true". (b) says "either P implies Q or P implies R" or "either "if P is true then Q is true" or "if P is true then R" is true".

Easiest way to show that they are not the same is to find a situation in which one is true and the other not- basically, write out the truth tables. Since each of P, Q, and R can be either True or False, there are 8 possiblities:
P Q R
T T T
T T F
T F T
T F F
F T T
F T F
F F T
F F F

Determine whether P |= (Q v R) is true or false in each of those cases, then determin whether (P |= Q) v (P|= R) is true of false in each of those cases. Can you find a case in which one is true but the other false?

 

[EvLer]

if i understand correctly, that does not seem to work out, since P |= (RvQ) is true if either R or Q are F but not both and so is the (P |= Q) v (P |= R), i.e. i do not see a case where one expression is true while the other is false...

 

[nonvestigial]

Qualitative explanation:

P entailing R means something in P MAKES R be true, same for entailing Q. Entailing both would entail P&Q. Maybe there's only something in P that narrows it down to Q or R, but doesn't force it either way, if you know what I mean?

Like the empty set entails P V ~P, but it doesn't entail P, it doesn't entail ~P, and it certainly doesn't entail both.