Conditionals: the material conditional

In this video, I examine the material analysis of indicative conditionals: the view that "if A then B" is true just in case A is false or B is true.
First, I give a prima facie problem for this analysis: the paradoxes of material implication. I then examine two attempts to defend the material analysis from this problem: Paul Grice's defence, which draws on his theory of conversational implicature; and Frank Jackson's, which draws on the theory conventional implicature and the concept of robustness. Finally I note some further problems that neither of these defences address.
For further reading, I strongly recommend Jonathan Bennett's "A Philosophical Guide to Conditionals", chapters 2 and 3. Stephen Read's "Thinking About Logic", chapter 3, is also a useful, though far less detailed, source.

[@IntegralDeLinha]

Awesome video! I love the crazy absurd examples!

[@ianpatrickmchugh787]

Brilliant! Thank you for posting this!

[@jamesclark1207]

This is a really fantastic video. Thank you very much for making it. I will be subscribing and watching your other videos.

[@RagingBlast2Fan]

Are you open to video suggestions? If yes, would you be willing to explore theory of justification or a particular theory of truth? Perhaps a critique of evidentialism, or something of that sort. Epistemology is a very interesting subject, but it seems you haven't really examined it very thoroughy in so far as your channel goes.

[@pareshhate2755]

Could you help me do this basic logic conditional proof sum using rules of inference and replacement?
1. F → G
2. J → ~G
/ to prove: F → ~J
3. F {C.P. Assumption}
4. G {1, 3, Modus Ponens}
I don't know what to do beyond this. :/

[@ninjakartezic5999]

1. F → G
2. J → ~G
3. F (Assumption)
4. J (Assumption in assumption)
5. G (MP)
6. ~G (MP)
7. ~J (RAA)
8. F → ~J (Introduction →)

[@TheYahmez]

1: F=>G
2: J=>¬G ∴F=>¬J
3 F (AFCP)
4 G 1,3:MT
5 ¬J 2,4:MP (CP)
6 ∴F=>¬J 3,5:Conditional Proof
---
(MP) Modus Ponens
1: A=>B
2: A
3 ∴B
(MT) Modus Tollens
1: A=>B
2: ¬B
3 ∴¬A
Proof By MP
1: A=>B
2: ¬B
3 ¬B=>¬A 1:Contraposition ( (A=B) (¬B=>¬A) )
4 ∴¬A 2,3:MP

[@miuhv1]

Another great video, but the jokes were sooo bad :D