Tutors Answer Your Questions about Proofs (FREE)
Question 1038664: I have bee having trouble solving these problems and been at it for hours I need help. Anyone's help would be greatly appreciated. I have to prove them.
Problem #1
1) (B^A)->D
2)A
3) C->~D/~(C^B)
4) :. (C^B)
Problem #2
1) (D^S)^T
2) [(Sv~F)vZ]->C/C^T
3) :. C^T
Problem #3
1) A->B
2) D->E
3) ~(B^E)/ ~Dv~A
4) :. ~Dv~A
Problem #4
1)~[(A^B) v (Cv~R)]
2) (T^~S)-> (Cv~R)
3) ~S/~T
4) ~T
Problem #5
1) (~Z^W)->Q
2) ~Z
3) R<-> (W^~Q)/~R
4) ~R
Click here to see answer by solver91311(24713)   |
Question 1039101: Hello, I need help with this Statement/Reasons type question.
Complete the following solution by filling in the missing statements and reasons.
•Statement: x-2(3x-4)=8 Reason:Given
•Statement:_________ Reason:Distributive Property
•Statement:-5x+8=18 Reason:Simplify
•Statement:_________ Reason:____________
•Statement: x=-2 Reason:____________
Thank you so much (:
Click here to see answer by jim_thompson5910(35256) |
Question 1041047: Note that ‘->’ is used for conditionals, ‘~’ is used for negations, ‘v’ is used for disjunctions, ‘&’ is used for conjunctions, ‘<->’ is used for biconditionals, and ‘/’ is used as the conclusion indicator.
Prove the following three arguments to be valid using the method of Natural Deduction. Do not use Conditional Proof in these proofs.
1. ~(P v Q)
2. (R v S) -> P / ~S
1. P v (Q v R)
2. (~T & ~R) / (Q v P)
1. (P v Q) -> R
2. (S v R) -> T
3. ~(Q -> E) / T
Click here to see answer by robertb(5830)  |
Question 1041143: Solve the following proof using the methods of natural deduction, as you did in the HW problems in Chapter 7. There may be many ways to solve the proof. You get partial credit for correct use of rules, even if you don't finish the proof. Incorrect use of rules with be penalized.
Good luck,
For the horseshoe use the > sign
For the Triple bar use the = sign
For the and use the . sign
For the or use the v sign
For negation use the - sign
1. (G . H) v (M . G)
2. G > (T . A) / A
Click here to see answer by solver91311(24713) |
Question 1042178: I have been super stuck on a logic puzzle and I need help! I have to translate it into symbolic form then use the rules, implication rules and replacement rules, along with indirect proof to derive the conclusion.
Here is the puzzle with the letters for the simple statements provided:
If the murder happened in the hotel room, then there are bloodstains somewhere in the room. It follows that it is not the case that the murder happened in the hotel room and there are not bloodstains somewhere in the room. (M, B).
Click here to see answer by solver91311(24713) |
Question 1044147: Can anyone tell me if I'm on the right track with this:
Question:
Let G = (V, E) be a directed graph. Let u, v ∈ V be distinct vertices. Define C(u) to be a
strongly connected component of G that contains u, and C(v) a strongly connected component
of G that contains v. Prove that either C(u) and C(v) do not share any vertex or C(u) and
C(v) are equal. [5 marks]
[Hint: To write such a proof you can begin with: “If C(u) and C(v) do not share any vertex
then the result is true, so suppose C(u) and C(v) share a vertex.”]
My answer:
A directed graph is strongly connected if there is a path from a to b, and a path from b to a, whenever a and b are vertices in the graph.
therefore if C(u) and c(V) share a vertex say w,
for the graph to be directed there must be strongly connected components between C(u) and C(w), and strongly connected components between C(v) and C(w)
Therefore as C(u) is strongly connected to C(w) and C(v) is strongly connected to C(w) (we have previously proved this theorem u,v,w E G(V,E) so c(u) is strongly connected to C(v) so therefore as they are both strongly connected to w in order for the definition to hold they must be equal and part of the same graph
Click here to see answer by richard1234(7193)  |
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