A -> C / (A^B) -> C
/ (A^B) -> C
I would think that it has to start with a conditional exchange, but have no idea where to start. There is a lot of reference to logic on the mises forums, so how about someone on the forums show me that they understand it.
I need help with this one too...
A -> ~A / ~A
/ ~A
I thought I was getting good at proofing, but not if there is only one premise.
This may help a bit:
1. A and B implies A. A implies C. therefore A and B implies C.
2. A imples not A [given].
not A implies not A [tautology]
Therefore A or not A implies not A. [Some law about or]
But A or not A is always the case [Law of excluded middle]
Therefore always the case that not A
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It's easy to refute an argument if you first misrepresent it. William Keizer
Thanks for responding.
What allows me to infer from conditionals? I am not allowed to use conditional proofsod indirect proofs for these two.
i got two from this. thanks!
If "A and B" is true then "A" is true and "B" is true (or whatever jargon you want to use) is called simplification or conjunctive elimination
A implies B
B implies C
Therefore A implies C.
You don't have that one in your repetoire?
We do not know that "A and B" is true. We have A implying C and that is it. If (A^B) was a premise I wouldn't have needed help.
A implies B B implies C Therefore A implies C. You don't have that one in your repetoire?
Well, you are trying to use Hypothetical Syllogism with two premises. My problem has only one premise and A implies B is not one of them. We do not know that B implies C. We are trying to prove that B doesn't interfere with A's implication of C.
The point of the exercises is to manipulate the inference rules and replacement rules.
You have A and B implies A, because that's a tautology.
You have A implies C
Thus A and B implies C. [by Hypothetical Syllogism with two premises.]
Of course, I don't know what the rules of the game are in your case, so it may not be what you are looking for.
It is not a premise. We do not have "A and B implies A" becase we do not have "A and B."
I am allowed to use any inference rule or any replacement rule. Excluded are ONLY CP and IP (it is the next part of the HW).
Have you ever done a proof?
It's been a while. I should think tautologies can be snuck in there.
Anyway, that's all I've got. Am interested in the answer, when you find it.
Unfortunately, the precise way of proving logical propositions like these aren't standardized. For your first, you can infer A from "A and B" from a boolean elimination rule called "conjuction elimination".
For your second one, I suppose you can do a proof by contradiction. Suppose A, then since A implies not A, we have A and not A, a contradiction. QED
Unfortunately, the precise way of proving logical propositions like these aren't standardized.
I think it is a good thing that there are many ways to do them. I am just not confident in the moves I make all the time.
For your first, you can infer A from "A and B" from a boolean elimination rule called "conjuction elimination".
Can I infer things from the conculsion? I thought I could merely rearrange them so that I can arrive at the conclusion on different terms.
I cannot use Conditional or Indirect proofs (assumptions) on these particular problems.
For the second one I got, simply enough...
1.) A -> ~A Pr / ~A 2.) ~A v ~A (Conditional Exchange; line 1) 3.) ~A (Duplication; line 2)
2.) ~A v ~A (Conditional Exchange; line 1)
3.) ~A (Duplication; line 2)
I think you should consult your instructor on these.
It's best to ask your instructor.
My instructor sucks. He is in his fifth (!!) year of the graduate program...it is only a three year graduate program in philosophy at my University...
I mean its true
A->C
(A^B)/A
(A^B)->C
But without inferring the logical equivalent of the statement of (A^B), which is what that conjunction rule does, idk how you'd even get to the form of the conclusion. You got me man
I don't know what to tell you since the terminology you're using is specific to your class. And your last proof is not logically coherent at all. Look at it carefully.
Here's a proof by cases.
Case 1: A
By the premise, A implies not A. Therefore, not A.
Case 2: not A
Then clearly, not A.
QED
I know it is true, you cannot proof invalid arguments. Here is a proof from later in the HW. It is supposed to be harder than the first one...
1.) (A -> B) -> ~(C -> D) Pr 2.) ~(A v F) Pr / ~(D v F) 3.) ~A ^ ~F (DeMorgan's Law; line 2) 4.) ~F (Simplification; line 3) - Here half of my conclusion has been proven. 5.) (A -> B) -> (~C ^ ~D) (Conditional Exchange; line 1) 6.) (~A v B) -> (~C ^ ~D) (Conditional Exchange; line 5) 7.) ~A (Simplification; line 3) 8.) (~C ^ ~D) (Dysjunctive Syllogism; lines 6 & 7) 9.) ~D (Simplification; line 8) 10.) ~D ^ ~F (Conjunction; lines 9 & 4) 11.) ~(D v F) (DeMorgan's Law; line 10) - Done.
2.) ~(A v F) Pr
/ ~(D v F)
3.) ~A ^ ~F (DeMorgan's Law; line 2)
4.) ~F (Simplification; line 3) - Here half of my conclusion has been proven.
5.) (A -> B) -> (~C ^ ~D) (Conditional Exchange; line 1)
6.) (~A v B) -> (~C ^ ~D) (Conditional Exchange; line 5)
7.) ~A (Simplification; line 3)
8.) (~C ^ ~D) (Dysjunctive Syllogism; lines 6 & 7)
9.) ~D (Simplification; line 8)
10.) ~D ^ ~F (Conjunction; lines 9 & 4)
11.) ~(D v F) (DeMorgan's Law; line 10) - Done.
And your last proof is not logically coherent at all. Look at it carefully.
It makes perfect sense. I added links so you can see why it makes sense.
If I am trying to prove ~A, then all I have to do is replace the conditional with a disjunct to present a choice of ~A or ~A. The implication is replaced by a new choice in the dysjunct.
Smiling Dave, Duplication is tautology. So, you were right about it fitting in there.
For the second one, your conditional statement is true. Make a truth table.
It only makes sense when A is False and ~A is True. Otherwise, you have contradictions and a case when the conditional is False.