MHB Proofing Implications Using Negation: P=>Q & ¬P=>Q

[aconti]

1st proof: ¬P=>Q |- PvQ

2nd proof: P=>Q |- ¬PvQ

I think negation introduction and negation elimination should be used, can you share any thoughts on how you would work out the above?

Thanks

 

[solakis1]

1st proof: ¬P=>Q |- PvQ

2nd proof: P=>Q |- ¬PvQ

I think negation introduction and negation elimination should be used, can you share any thoughts on how you would work out the above?

Thanks

For the 1st one:

Proof :

1) ~P=>Q.....................Given

2) ~(PVQ)..................Assumption to lead to a contradiction

3) ~P & ~Q...................2, D.Morgan

4) ~P....................3, Addition Elimination

5) ~Q ....................3,Addition Elimination

6) Q.....................1,4 M.Ponens

7) Q&~Q ...................5,6 Addition Introduction

8) PVQ....................2 to 7 and Contradiction

Can you do the 2nd exescise also by contradiction ??