[TechSyndrome]

Hi. I've come across this link on YouTube that gives an example of Universal Generalisation in Formal Proofs of Validity:



But, I'm having a problem with line 8: Dr Jason Campbell uses material implication on line 7 of the second column - the problem I have with this is that the rules of inference says:



From what I'm seeing in the above rule there is no formula which has a direct correspondence to the formula on line 7 i.e. a negation on each side of the variable (I think that's what it's called). There is a negation only on the left variable of the rule, but not on the right. So, how did he still use the rule?

This post has been edited by TechSyndrome: 14 July 2012 - 03:07 PM

[sepp2k]

If you let P=Ay and Q=~Cy, you get the step that he performed from line 7 to 8. Again the variables in the rule can stand for arbitrary terms, not just single variable. So it's okay for Q to stand for ~Cy.

[TechSyndrome]

sepp2k, on 14 July 2012 - 03:22 PM, said:

If you let P=Ay and Q=~Cy, you get the step that he performed from line 7 to 8. Again the variables in the rule can stand for arbitrary terms, not just single variable. So it's okay for Q to stand for ~Cy.

Sorry if I misunderstood - when you say "P=Ay and Q=~C" which side of the rule are you looking at?

Btw, I hope you don't think "I'm dealing with an idiot". There are some things that I just instantly get, and am able to simplify, while there are some things that I take some time to understand.

Edit: And I appreciate your help - a LOT.

Edit Edit: And I go through these eureka moments where I'm like "OMG so that's what you mean!" and I get crazy excited over it.

This post has been edited by TechSyndrome: 14 July 2012 - 04:09 PM

[sepp2k]

I'm looking at all sides of the rule. If a variable means one thing on one side of the rule, it must mean the exact same thing on the other side of the rule.

I'm saying that if we substitute Ay for P and ~Cy for Q, then the rule "P -> Q <-> ~P v Q" becomes "Ay -> ~Cy <-> ~Ay v ~Cy" and now the two sides of the <-> operator are exactly line 7 and 8 of the picture you posted.