propositional proof

[chrislav]

Given the following equivalence :

[(p^q)=>r] <=>[p=>(q=>r)]

Can we prove it formaly without using the law of conditional proof (deduction theorem)?

 

[Steven G]

Try it yourself and post back your work

 

[logistic_guy]

Try it yourself and post back your work

I have been on this forum for just over 10 years and have decided to not be a helper anymore as long as this logistic_guy is still around. I enjoyed my time here, but as I said it's time to say goodbye.
Steven

You are a liar as I am still here.

 

[chrislav]

i think that there is not another solution except that where we use the conditional proof.
I tried.

 

[blamocur]

I can't think of a simpler solution than building truth tables for both sides.

 

[Aion]

Given the following equivalence :

[(p^q)=>r] <=>[p=>(q=>r)]

Can we prove it formaly without using the law of conditional proof (deduction theorem)?

You can also prove it purely by equivalence transformations (no conditional proof). Use only the truth-functional laws for [imath]\to,\neg,\land,\lor[/imath].

 

[Agent Smith]

It's an equivalence rule in natural deduction.
If you kiss the girl and she smiles then that's love <=> If you kiss the girl then if the girl smiles then that's love.

Intuitive proof