Formal Logic Question Forum

[JMCSpartan08]

Can you combine two "All" conditionals, to get a valid conclusion, that look like this:

A->C
B->C
----
A->B

OR

A->B
A->C
----
B->C


My gut tells me "no" but I wanted to double check.

[Malapropism]

Nope. You can only combine like

A-->B
B-->C

to A-->C

Think of it like this:
For the first example, if A is cats, and B is dogs, and C is are animals, all cats are animals, all dogs are animals, but all cats are not dogs.

For the second, if A is cats, B is are animals, and C is have teeth, all cats are animals, all cats have claws, but all animals don't have claws.

I hope that clears things up

[poppy]

In the first one you get:

A----> C <-----B

from this you can't infer anything because in order to connect all the variables you'd have to change it to be:

A <-----> C <------> B the double arrow meaning some, and with two some double arrows you can't make an inference

On the second one you get:

B <----- A ------> C

This one you can do B <------> A ------> C or B<------- A <--------> C Both mean the same thing. And from one Some arrow and an All arrow pointing away from the Some arrow you can make inferences.

From this you can infer B <--------> C some B's are C's

[boblawlob]

In the first one you get:

A----> C <-----B

from this you can't infer anything because in order to connect all the variables you'd have to change it to be:

A <-----> C <------> B the double arrow meaning some, and with two some double arrows you can't make an inference

On the second one you get:

B <----- A ------> C

This one you can do B <------> A ------> C or B<------- A <--------> C Both mean the same thing. And from one Some arrow and an All arrow pointing away from the Some arrow you can make inferences.

From this you can infer B <--------> C some B's are C's

Can't you infer from the first statement that some not As are not Bs?

A->C (not C -> not A)
B->C (not C -> not B)

[bp shinners]

Can't you infer from the first statement that some not As are not Bs?

A->C (not C -> not A)
B->C (not C -> not B)

Yes, but you're never going to have to on the LSAT.