Logic 101 (#42): Nested Proofs

Proofs by contradiction and conditional proofs do not need to be done one at a time. Instead, you can layer them as necessary to prove what you are looking for. This video gives an example of a nested proof and discusses some of the rules for them.

[mxlexrd]

Here's a way to do the inner part using a conditional proof.
Q^~Q Assumption
Q Simplification
~Q Simplification
~QvR Disjunction introduction
Q=>R Material implication
R Modus ponens
In fact, what this shows is that a contradiction implies anything, since we started with a contradiction (Q^~Q) and derived an arbitrary conclusion (R).

[torosalvajebcn]

On line 3, should not "Conditional Exchange" be called Material Implication?

[PunmasterSTP]

Nested proof? More like "Nice, this is good!" Thanks again for these wonderful videos. Sometimes I get a bit down when I can see the end of the playlist coming into view, but I can't wait to start my next adventure!

[dinghaoluo]

but according to reductive explosion you can simply deduce R from the presence of Q ^ ~Q since it is already a contradiction is it not?

[bogdanoiul]

Looking forward to your next courses! You are an excellent teacher!
KUDOS.

[SchoolRumbleRawks]

I dont really understand how he got from step 4 to 5, how did the ~R disappear?
Which Law is this?

[Velzen]

if Q and not Q is a contradiction then this whole exercise seems nonsense to me,

[TheLukeskywalker]

A vacuous proof could help this one. It would still need to be nested though.

[lea]

I solved this the following way:

(Q and -Q) is always false so the conditional (Q and -Q)=>R is always true. So because that conditional is true, and is the consequent of the main conditional statement we are trying to prove here, we can take the main statement as proved too since the conjunction {P and -[(Q and -Q)=>R]} will always be false so - {P and -[(Q and -Q)=>R]}. Then I re-write {-P or [(Q and -Q)=>R]} with De Morgan. Then I do material implication to get the affirmation P=>[(Q and -Q)=>R].

This is something that came to my mind and I don't have experience besides watching this videos. Please if someone with more knowledge could tell me if this is right or wrong I'd appreciate.-

[DYNAMIN]

Sorry... This whole thing with taotologies confuses me. Wouldnt I be able to substitute P, Q and R for anny simple sentence I want. For example: if P is: space is empty, and R is: god exists. Does that meen that god exist or whatever you say R to be is true?

[torosalvajebcn]

Yo forgot to close the conditional proof.

[XfireSSBU]

I used a second conditional proof.
2 Q ^ ~Q Assumption
3 ~Q Simplification
4 Q Simplification
5 ~Q v R Disjunction intro
6 Q => R Material implication
7 R Modus Ponens
Is this a valid argument?