There looks to be an error in the first worked tutorial example - see below Given 1. (p⊃q)·(q⊃~p) 2. p∨q 3. ~q / s·t The proof as given in the tutorial 1. (p⊃q)·(q⊃~p) 2. p∨q ⇒ (q∨~p) ... (4) 4. (q∨~p) 3. ~q ⇒ ~p ... (5) 5. ~p 2. p∨q ⇒ q ... (6) ⇒ q ∨ (s·t) ... (7) 3. ~q 7. q∨(s·t) ⇒ s.t =========================== Now, from (3) and (6) above we have 3. ~q is true 6. q is true ⇒ ~q = q How can this be possible?
Thanks! I'll go back and double check when I get a chance. It's entirely possible that there's a mistake. Super busy with the end of the semester, but I'll definitely get to it.
![BLACK BAG - Official Trailer [HD] - Only in Theaters March 14](http://i.ytimg.com/vi/Du0Xp8WX_7I/mqdefault.jpg)
Thank you so much. This video was Incredibly helpful!
Thanks so very much! I made them for the students I teach, but decided to share them with everyone.
There looks to be an error in the first worked tutorial example - see below
Given
1. (p⊃q)·(q⊃~p)
2. p∨q
3. ~q / s·t
The proof as given in the tutorial
1. (p⊃q)·(q⊃~p)
2. p∨q ⇒ (q∨~p) ... (4)
4. (q∨~p)
3. ~q ⇒ ~p ... (5)
5. ~p
2. p∨q ⇒ q ... (6) ⇒ q ∨ (s·t) ... (7)
3. ~q
7. q∨(s·t) ⇒ s.t
===========================
Now, from (3) and (6) above we have
3. ~q is true
6. q is true ⇒ ~q = q
How can this be possible?
Thanks! I'll go back and double check when I get a chance. It's entirely possible that there's a mistake. Super busy with the end of the semester, but I'll definitely get to it.
Hi dr Barry, did you get a chance to check the error ?