SOLUTION: 1. P v R
2. R -> Q ////// :. (P v Q)
Algebra.Com
Question 973154: 1. P v R
2. R -> Q ////// :. (P v Q)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Prove by exhaustive cases:
Since P v R is given, the case P and case R are exhaustive.
Assume P. Then P v Q by Disjunction Introduction.
Assume R. Then Q from R -> Q by Modus Ponens and then P v Q by Disjuction Introduction.
John

My calculator said it, I believe it, that settles it
RELATED QUESTIONS
P v (Q & R) <=> (P v Q) & (P v R)
(answered by Edwin McCravy)
1. P V Q
2. Q -> (R & S)
3. (R V P) -> T /... (answered by Edwin McCravy)
~p ^ r
p v q
--------
∴ q v... (answered by math_tutor2020)
(q →... (answered by MathLover1)
1. ~(U v R)
2. (~R v N) ⊃ (P * H)
3. Q ⊃ ~H /... (answered by math_helper)
1. ~(U v R)
2. (~R v N) ⊃ (P • H)
3. Q ⊃ ~H /~Q
(answered by CPhill)
I need help finishing this proof.
(P . Q) -> (R v S) :. (P -> R) v (Q -> S)
1. (P . (answered by jim_thompson5910)
I need help finishing this proof.
(P . Q) -> (R v S) :. (P -> R) v (Q -> S)
1. (P . (answered by jim_thompson5910)
Using Inference Rules
Premise 1. (P * Q) > R
Premise 2. (P * Q) V S
Premise 3. ~ R (answered by Edwin McCravy)