SOLUTION: I have been wrestling with trying to solve this proof.....Pleeaaassseee HELP! (J->L)->(M->N),A,A->(J->L),(M->N)->[(J->L)->O],O->(A->B)|-B

Algebra ->  Proofs -> SOLUTION: I have been wrestling with trying to solve this proof.....Pleeaaassseee HELP! (J->L)->(M->N),A,A->(J->L),(M->N)->[(J->L)->O],O->(A->B)|-B       Log On


   

Question 707955: I have been wrestling with trying to solve this proof.....Pleeaaassseee HELP!
(J->L)->(M->N),A,A->(J->L),(M->N)->[(J->L)->O],O->(A->B)|-B

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You have A (ie you're assuming A is true) and you have this

A -> (J -> L)

So you can use modus ponens to get J -> L

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You have this statement

(J->L)->(M->N)

and you can use modus ponens with this and the "J -> L" you got earlier to get M -> N

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Now you have M -> N and (M->N)->[(J->L)->O]

Use modus ponens to get (J->L)->O

Then use modus ponens again with J -> L and (J->L)->O to get O

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You now have O and O -> (A -> B)

Use modus ponens to get A -> B

Finally, use modus ponens with A and A -> B to get B

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So all of the given statements and using modus ponens a bunch of times leads us to the conclusion of B

This proves that the argument and conclusion is valid.