0

This is where I’ve gotten so far. I’ve proven it from left to right but I’m getting some trouble proving it from right to left. I’m trying to reach the conclusion by double negation.

My Progress

  • See Rule of Material Implication. – Mauro ALLEGRANZA Oct 9 '19 at 18:38
  • My professor only wants us to use the introduction and elimination rules – Sanjeev Oct 9 '19 at 20:04
  • 1
    You might be able to show this directly by assuming A on line 2 and then considering the two cases in line 1. In both you will need to derive B. Here is a proof checker you can use to guide your work: proofs.openlogicproject.org – Frank Hubeny Oct 9 '19 at 20:20
  • 1
    Well you can use conditional proof with assuming $A$, and then double negation of $A$, and then destructive syllogism on ~AvB and ~~A, to deduce B. – MathematicalPhysicist Oct 9 '19 at 20:31
  • Have you completed this yet? – Graham Kemp Nov 9 '19 at 8:10
0

I’m trying to reach the conclusion by double negation.

Well, you might do that, but it is not indicated. You have a premise that is a disjunction and a conclusion that is a conditional.

You will want to use a Conditional Proof to introduce the conditional, so assume A attempting to derive B.

You will want to use a Proof by Cases to eliminate the disjunction in the premise. So do that .

   1 (1) ~A v B  Premise
   2 (2) A       Assumption
   3 (3) ~A      Assumption
 ... (4) ...     ...
 2,3 (5) B       ...
   4 (6) B       Assumption
 1,2 (7) B       Disjunction Elimination 1,2-5,6-6
   1 (8) A -> B  Conditional Introduction 2-7
| improve this answer | |
0

I took symbolic logic a while ago, but I believe this should be correct. Apologies for my sometimes messy handwriting. proof

| improve this answer | |
New contributor
A.O. is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
-1

The truth table definition of A -> B says that:

  • ~A implies A -> B is true
  • B implies A -> B is true

enter image description here

PROOF OF (~A v B) --> (A --> B)

0 ....    (~A v B)         ..... assumption         
1 ....    ~A --> (A --> B) ..... by truth-table definition of logical implication          
2 ....    B  --> (A --> B) ..... by truth-table definition of logical  implication        
3 ....    A --> B          ..... 0, 1, 2  proof by cases
| improve this answer | |
  • This is not a natural deduction proof. – lemontree Apr 4 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.