Mathematica Mathematical Logic: For all and There exists

Click For Summary
The discussion revolves around the logical expression involving universal quantifiers, specifically the implications of the notation "for all" (∀) and "there exists" (∃) in mathematical logic. The primary focus is on proving the equivalence of two statements involving these quantifiers. The original poster expresses confusion about how the presence of the universal quantifier (∀x) affects the logical structure of the equation and whether it can be ignored for simplification. Participants clarify that the symbols "→" denote implication and that the presence of quantifiers is crucial in determining the validity of logical statements. The deduction theorem is mentioned as a method for manipulating the expression, but concerns are raised about whether traditional methods apply when quantifiers are involved. Overall, the discussion emphasizes the importance of understanding the role of quantifiers in logical expressions and their impact on proofs.
Goldenwind
Messages
145
Reaction score
0
Mathematical Logic: "For all" and "There exists"

I need to show that
\vdash (\forall x)(A \rightarrow (B \equiv C)) \rightarrow ((\forall x)(A \rightarrow B) \equiv (\forall x)(A \rightarrow C))

My question to you, how does the (\forall x) affect this equation? If they weren't there, I could simply do this question, but their presence is confusing me. What's different? Can I just ignore them and move on as normal?
 
Last edited:
Physics news on Phys.org
Are your "\rightarrow" symbols for "implies"? If so, use \Rightarrow next time, so it isn't so confusing.

Now, with the notation you chose, I don't really understand what the entire statement is conjecturing. Please explain (in words) what you are trying to ask. (Besides the question about \forall x.)
 
I wrote the symbols exactly how I was taught, sorry =/
I was taught that \rightarrow is used for "implies", or (¬A v B).

Now for what I'm trying to show... See the 2nd "implies" symbol? I'll be using the deduction theorem to move the (Ax)(A --> (B = C)) over to the left side of the |--, and then will attempt to work with the remaining right side of the |-- to show that it can be expressed the same as the left.

The thing is, my methods work when (Ax) isn't there, however I'm not sure if they work the same when it is. Can I just ignore the presence of (Ax), and do this question as if it weren't there?
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K