Subscription pricing
Question 1:
For premise: A --> B
Are all of the following valid inferences?
A -m-> B
A -s- B
-B --> -A
-B -m-> -A
-B -s- -A
Question 2:
For premise: A --> B --> C
Is the following necessarily false?
A --> -C
Question 3:
From premise A --> B --> C is there any inference we can make that is necessarily false?
0
8 comments
@businesskarafa858 said:
I thought that if the premises are true then a valid conclusion is necessarily true and never false. Is that the case?
True Premise1: A --> B
True Premise2: A --> -B
True Premise3: A
Valid/True Inference: A --> B and -B
Valid/True Inference: A and B and -B (because the A was 'activated' by true premise 3
Sorry if it seems like I side stepped what you were explaining previously, I have a few outlying concepts in LR that I'm not sufficiently comfortable with and I'm trying to tie them together.
Definitely, if the premises are true than all valid conclusions are true.
Each inference in the example above is valid, but they also result in an impossible paradox: "B and B." It's the same as saying you do have a dog and also you don't, or that you exist and don't exist.
Since the inferences are valid and the conclusion is impossible, the only thing left is to attack the premises. At least one of the three premises has to be untrue- no set of true premises will result in an impossible outcome.
@joelatennyson369 said:
In both abstract and concrete form you’d just call that a contradiction on LR. It’s probably the literal definition of a contradiction. I feel like I’ve seen it before somewhere but it’s presumably most expected in a flaw Q prob.
Again, thanks for the responses. My understanding has definitely increased :smile:
I thought that if the premises are true then a valid conclusion is necessarily true and never false. Is that the case?
True Premise1: A --> B
True Premise2: A --> -B
True Premise3: A
Valid/True Inference: A --> B and -B
Valid/True Inference: A and B and -B (because the A was 'activated' by true premise 3
Sorry if it seems like I side stepped what you were explaining previously, I have a few outlying concepts in LR that I'm not sufficiently comfortable with and I'm trying to tie them together.
Definitely, if the premises are true than all valid conclusions are true.
Each inference in the example above is valid, but they also result in an impossible paradox: "B and B." It's the same as saying you do have a dog and also you don't, or that you exist and don't exist.
Since the inferences are valid and the conclusion is impossible, the only thing left is to attack the premises. At least one of the three premises has to be untrue- no set of true premises will result in an impossible outcome.
In both abstract and concrete form you’d just call that a contradiction on LR. It’s probably the literal definition of a contradiction. I feel like I’ve seen it before somewhere but it’s presumably most expected in a flaw Q prob.
*mistake
@businesskarafa858 said:
Technically yes, that's valid.
It's impossible for it to be true that 'B and B.' So if you know that A --> B and B, then you can actually be certain that A.
There's a relevant logic game, PT34 S4 G4. It's the only time I've ever seen this inference being important on the actual LSAT. Spoilers for the game below:
! When you string all the conditionals together, one of the findings is that L --> O and O. The key inference is that L must therefore never be selected (selected = Souderton, not selected = Randsbourough). It's such a weird game because when you see a contradiction like that you'd usually assume that you wrote something down wrong.
I thought that if the premises are true then a valid conclusion is necessarily true and never false. Is that the case?
True Premise1: A --> B
True Premise2: A --> -B
True Premise3: A
Valid/True Inference: A --> B and -B
Valid/True Inference: A and B and -B (because the A was 'activated' by true premise 3
Sorry if it seems like I side stepped what you were explaining previously, I have a few outlying concepts in LR that I'm not sufficiently comfortable with and I'm trying to tie them together.
Technically yes, that's valid.
It's impossible for it to be true that 'B and B.' So if you know that A --> B and B, then you can actually be certain that A.
There's a relevant logic game, PT34 S4 G4. It's the only time I've ever seen this inference being important on the actual LSAT. Spoilers for the game below:
! When you string all the conditionals together, one of the findings is that L --> O and O. The key inference is that L must therefore never be selected (selected = Souderton, not selected = Randsbourough). It's such a weird game because when you see a contradiction like that you'd usually assume that you wrote something down wrong.
Thanks for the response. I think I understand but I have another question:
Premise: A --> B
Premise: A --> -B
Inference: A --> B and -B
Is this inference valid?
Q1) All look good to me. Valid.
Q2) Yes, it's false.
Q3) See Q2.
Great :)