How to prove H → M ￢H → ￢M prove H↔M?

Question

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?

Answer 1

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.

Answer 2

????

(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)

Answer 3

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.