P=>q, q=>r, then p=>r (proof assumes p why?)

  • Thread starter chisser98
  • Start date
  • #1
2
0
hi guys,

I'm struggling to learn logic and I'm stuck on what I'm sure is an easy answer. Here's the question:

whenever p is true, q is true. Whenever q is true, r is true. Prove that, whenever p is true, r is true.

Here's part of the answer:

p=>q Premise
q=>r Premise
(q=>r) => (p => (q=>r)) Implication Introduction

...etc

I can follow the proof just fine from this point. However, why can we assume p is true? The Implication Introduction step is allowing us to assume p is true..even when it's not a premise (or at least they don't state it as a premise explicitly).

Anyway, like I said, I'm sure this is easy, I'm just not wrapping my head around it.

Any help would be appreciated. Thanks guys!
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
The Implication Introduction step is allowing us to assume p is true..
No it's not. S is not a consequence of S => T.
 
  • #3
2
0
Thanks for the reply Hurkyl. Ok - since S is not a consequence of S=>T, we can say it's true? I'm saying this because the implication introduction scheme is:
M => (N => M)

in this example, we've set M = (q=>r) and N = p, so we get:
(q=>r) => (p => (q=>r))

I guess I'm just not understanding why we can assign N=p when p isn't an explicit premise?
 

Related Threads on P=>q, q=>r, then p=>r (proof assumes p why?)

  • Last Post
Replies
6
Views
17K
Replies
12
Views
5K
Replies
3
Views
12K
Replies
2
Views
11K
  • Last Post
Replies
5
Views
7K
Replies
3
Views
720
Replies
8
Views
4K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
780
  • Last Post
Replies
2
Views
571
Top