-1

I'm using the program Fitch and I need to make a formal proof for this:

  1. H → M
  2. ¬H → ¬M

Prove: H↔M

Any ideas on how to do so?

  • Welcome to SE Philosophy! Thanks for your contribution. Please take a quick moment to take the tour or find help. You can perform searches here or seek additional clarification at the meta site. Don't forget, when someone has answered your question, you can click on the checkmark to reward the user! – J D Nov 25 '19 at 0:42
  • Hello, it might be difficult to get homework help unless you shown you have put some effort into it and tried some things. Here's a good example.. – J D Nov 25 '19 at 0:45
  • See also this post – Mauro ALLEGRANZA Nov 25 '19 at 10:58
  • what is an idea? sorry, you're using language in a highly imprecise way here – user38026 Dec 27 '19 at 3:31
0

Okay,

Normally I keep my nose out of logic especially, but this one is straightforward, so I'll give you a clue.

Note, that in this argument, you have a conclusion which is the biconditional. The biconditional's definition requires two criteria be met, and ONE of the two premises in the argument satisfies it. The other doesn't, but I'm suggesting you might want to write out the inverse, converse, and contrapositive of both premises and see how they relate to the definition of the biconditional.

Once you find the two premises to satisfy the definition of the biconditional, you should be all set!

Good luck.

| improve this answer | |
  • I think it would be best at this point to just give your answer. – Eodnhoj7 Nov 27 '19 at 1:03
0

????

(H --> H) --> M

H --> M

(H --> -H) <--> (M <--> -H)

(-H --> -H) --> --H

-H --> H

(-H --> -M) <--> (-M <--> H)

(H --> M --> -H --> H --> -M) --> (H <--> M)

| improve this answer | |
  • What does "????" mean? – Mark Andrews Nov 27 '19 at 1:20
  • Thought I am not entirely sure about, as in "question it". – Eodnhoj7 Nov 27 '19 at 1:23
0

You have H → M as one premise, so deriving M → H will allow you to introduce the biconditional. So introduce that conditional the usual way (aka via a conditional proof).

|  H → M       Premise
|_ ¬H → ¬M     Premise
|  |_ M        Assumption
|  |  :
|  |  H
|  M → H      Conditional Introduction
|  H ↔ M      Biconditional Introduction

The steps between the assumption of M and the derivation of H should not be hard. Looking at the second premise will be helpful.

| improve this answer | |
  • How is H not also an assumption? – Eodnhoj7 Nov 27 '19 at 0:11
  • Because you are not assuming H, you are aiming to derive it from the assumption of M. However you can do that by making another assumption – Graham Kemp Nov 27 '19 at 0:13
  • 1
    We are not assuming H anywhere! No subproof has been raised with an assumption of H. – Graham Kemp Nov 27 '19 at 1:06
  • 1
    H is the antecedent of the conditional statement H → M. The words are not synonyms. – Graham Kemp Nov 27 '19 at 1:14
  • 2
    In the Fitch System of Natural Deduction an assumption is a statement raised (without needing to be derived) to start a context of contingent derivations (aka a subproof). That is all that it is. – Graham Kemp Nov 27 '19 at 2:39

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.