Logic 101 (#18): Contraposition

Fun fact: p (arrow) q is the same thing as ~q (arrow) ~p. This lecture explains why using neat pictures.

[julesjgreig]

Very clear, William - thank you for sharing this.

[PunmasterSTP]

Contraposition? More like “contraption”…for learning! These videos are an amazing creation; keep up the good work!

[TheMorhaGroup]

IF,
[P => Q] <=> [-Q => -P]

AND,
[Q => P] <=> [ -P => -Q]

WHERE,
[-Q => -P] <=> [-P => -Q] = ∆
[Such equivalence is drawn on the basis of your set example. Where not being in Q, the main set in which P is a subset of, means I cannot be in P, as one must first be in Q, to be able to be in P. And If I am not in P, still a subset of Q, then I can not be in the main set Q, as one must first be in Q, to be able to be in P, just as before]

THEN,
[P => Q]=∆,
[Q => P]=∆
[P => Q] = [Q => P]
Through their equivalence with ∆
Surely this can not be the case?

[nickmillican]

From 'Material Implication' and 'Contraposition', start with (-P v Q) <=> (P => Q) <= (-Q => -P). I take this to mean that (-P v Q) <=> (-Q => -P). But I don't think this holds...keep reading below.

P = I'm in the inner sanctum
Q = I'm in the Castle
Hence, P => Q <=> -Q => -P. If I'm in the inner sanctum, I'm in the castle; if I'm not in the castle, I'm not in the inner sanctum. Yes?
But then how can it follow that P=> Q <=> -P v Q as well? I'm not in the inner sanctum or I'm in the castle.

Going the other way around:
P = I am hungry
Q = I eat
Hence, P => Q <=> -P v Q. If I am hungry, then I eat; I cannot be hungry and not eat, though I may eat even though I'm not hungry. Yes?
But then how can it follow that P => Q <=> -Q => -P as well? I don't eat if I'm not hungry, even though -P v Q states that I may eat even if I'm not hungry.

Sure enough, you run it through the truth table and--so long as you take vacuously true statements to be true true statements, they are equivalent. But, if I'm reasoning clearly, they can't be equivalent.
Something that occurs to me, though I haven't thought it all the way the through, is that -Q => P seems to be descriptive (this is how it is) whereas -P v Q seems prescriptive (I'm asserting this).

Where am I going wrong?

[sarahlockhart-ragland]

Hello William I am taking a logic class and trying to figure all this out this is one of my questions
1. (S > B) > (S v K)
2. (K v ~D) > (H > S)
3. ~S . W / ~H

I am unsure on how to figure this out can you assist

[royhamilton]

Contraposition is the WRONG NAME. The legit name is Material TRANSPOSITION. Contraposition goes back to Aristotle and he did not have symbols. Contraposition further is not VALID for propositions of the form NO S is P and Aristotle pointed this out over 2, 000 years ago. So classical contraposition is only valid some times which only applies to two proposition forms which are A type propositions and O type propositions. I have no idea why mathematicians call this contraposition. There is no reason to rename stuff that pre existed. Why not redefine triangle while we are just messing stuff up.

[vle]

I personally find it much easier to use Boolean logic proofs.

[sarahlockhart-ragland]

Looking for the correct inference and have no clue on how to understand which one out of the 17 that this would fall in