Tutors Answer Your Questions about Proofs (FREE)
Question 928249: I am having trouble solving this proof:
1. DvM
2. M-->R
3. Sv(Q&X)
4. R-->~S
5. ~D /Q
So far I have:
6. M
7. R
8. (M&R)
Is this correct so far? This is the farthest I have gotten and don't quite understand how to solve the rest of this proof. Any help would be greatly appreciated. Thank you!
Click here to see answer by AnlytcPhil(1806)  |
Question 928647: I need to somehow get "R ≡ ∼Q" as the conclusion, using natural deduction (This means the 18 rules of natural deduction, Modus Ponens, Modus Tullens, De Morgan's Rule etc etc.)
Conclusion: R ≡ ∼Q
1. A ⊃ (Q ∨ R)
2. (R • Q) ⊃ B
3. A • ∼B
Click here to see answer by Edwin McCravy(20055)  |
Question 929923: Can anyone annotate the following proofs for me?
Proof 1:
1. (AvB) -> C
2. (CvD) -> E
3. CvA
4. ~C .: E
5. A
6. AvB
7. C
8. CvD
9. E
Proof 2
1. X->Y
2. X->Z .: X-> (Y&Z)
3. ~XvY
4. ~XvZ
5. (~XvY) & (~XvZ)
6. ~Xv(Y&Z)
7. X-> (Y&Z)
Proof 3
1. ~(A&(BvC)) .: A -> ~B
2. ~((A&B)v(A&C))
3. ~(A&B) & ~(A&C)
4. ~(A&B)
5. ~A v ~B
6. A -> ~B
PLEASE and thank you!!!! I need lines and rules.
Click here to see answer by Edwin McCravy(20055)  |
Question 932733: Can someone please help me with this proof using the 18 rules of inference?
> is a horshoe, v is the wedge, * is the dot
1. (S v T) > (S > ~T)
2. (S > ~T) > (T > K)
3. S v T Therefore: S v K
4. S > ~T 1,3 Modus Ponens
5. T > K 2,4 Modus Ponens
6. (S > ~T) * (T > K) 4,5 Conjunction
7. ~T v K 3,6 Constructive Dilemma
8. ~S v ~T 4, Material Implication
9. ~(S*T) 8, DeMorgan's Rule
I have been stuck on this question for hours, and I can't seek to get to S v K.
Click here to see answer by jim_thompson5910(35256) |
Question 935274: This is a Number Theory proof:
If N = abc + 1, prove that (N, a) = (N, b) = (N, c) = 1.
(N, a) means "the greatest common divisor of N and a."
I have started the proof like this:
Let d = (N, a).
Since d = (N, a), then d | N and d | a. "d divides N" and "d divides a".
Click here to see answer by KMST(5328)  |
Question 935397: 1. ~P∨Q
2. Q→R ∴ P→R
1. R∨~Q
2. P→Q
3. ~R ∴ ~P
1. P
2. ~Q
3. ~R
4. (P&~Q)→(R∨S) ∴ S
1. (P&Q)→R
2. (P&R)→Q
3. P ∴ R↔Q
1. ~P&~Q
2. (P∨Q)∨(R∨S) ∴ (R∨S)
1. ~P ∴ ~(P&Q)
1. (P→Q)→S
2. S→~(T∨W)
3. (~T&~W)→R ∴ (P→Q)→R
1. (P&~Q)&(~R∨S)
2. ~((S&P)&~Q) ∴ ~R
1. P→Q
2. P→~(R∨Q) ∴ ~P
10. ∴ (P→~Q)↔(Q→~P)
Click here to see answer by Edwin McCravy(20055)  |
Question 943566: A donkey must transport 900 carrots to the market, which is 300 miles away. The donkey carries a maximum of 300 carrots, and eats 1 carrot every mile. What is the largest number of carrots that can be delivered to the market?
Click here to see answer by macston(5194)  |
Question 943424: Prove (m+n)(p+q) = (mp+np) + (mq+nq)
I have already proved a proposition that states (m+n)p = mp + np, which matches up with part of the right side of the equation, but I'm not sure how to get to that point or how to incorporate mq+nq into it.
Click here to see answer by richard1234(7193)  |
Question 957379: Boolean logic is the form of algebra in which all values are needed to either TRUE or FALSE normally through AND,OR and NOT operator.
a)How do the number of result change upon using an "AND" operator and an "OR" operator?EXPLAIN your answer with the help of relevant examples.
b)What is the name for the combination logic gate of an "AND" operator followed by a "NOT" operator?Explain the whole operation and provide the reasons on why do we need this combination.
Click here to see answer by Edwin McCravy(20055)  |
Question 959729: Some things are caused to exist.
Anything caused to exist is caused by another [Nothing can bring itself into existence].
If some things are caused to exist and anything caused to exist is caused by another, then
either there’s a uncaused first cause or there’s an infinite series of past causes.
There’s no infinite series of past causes.
There’s an uncaused first cause.
Click here to see answer by Alan3354(69443)  |
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