Why Does the Outcome Change When Solving pvq->r Differently?

[TheNaturalStep]

Sorry if my title wasn't the best, doesn't know the english termonolgy

Anyway i get diffrent result based on how i solve "pvq->r" the result vary if i solve this as (pVq)->r or pV(q->r).

You can see me trying to solve this in the picture which i have attaced, below

http://img183.imageshack.us/img183/2387/dscf0462ho0.jpg

Thank you for your time

 

[HallsofIvy]

Sorry if my title wasn't the best, doesn't know the english termonolgy

Anyway i get diffrent result based on how i solve "pvq->r" the result vary if i solve this as (pVq)->r or pV(q->r).

You can see me trying to solve this in the picture which i have attaced, below

http://img183.imageshack.us/img183/2387/dscf0462ho0.jpg

Thank you for your time

Yes, you should- those have different meanings: (pvq)-> r means "if either p or q is true, then r is true." pv(q->r) means "either p is true or if q is true then r is true". The "precedence" rules for symbolic logic say that "and", v, takes precedence of "implication, ->, just as in arithmetic multiplication takes precedence of over multiplication takes precedence over addition: ab+ c means (ab)+ c. pvq->r, without parentheses, means (pvq)-> r.