Show that the argument (p∧q)∨(¬p∧¬q)⊢p⊃q is valid with natural deduction

Question

Here is what I have done :

 1  | (p∧q)∨(¬p∧¬q)       H 
 2  |  | (p∧q)            H
 3. |  | | p              H
 4. |  | | p∧q            R-2
 5. |  | | q              ∧E-4
 6  |  | p⊃q 

I'm now trying to demonstrate that (¬p∧¬q) can also imply p⊃q so that I can conclude p⊃q with de rule ∨E. I've tried many different things but can't manage to get the result. Any help is appreciated.

Answer 1

I'm now trying to demonstrate that (¬p∧¬q) can also imply p⊃q so that I can conclude p⊃q with de rule ∨E.

The principle of explosion, or ex falso quadlibet, says that you may derive any statement from a contradiction.

Since you can derive a contradiction under the assumptions of ¬p∧¬q and p, then you may deriver q.

 1  | (p∧q)∨(¬p∧¬q)        H 
 2  |  |_ (p∧q)            H
 3. |  |  |_ p             H
 4. |  |  | p∧q            R-2
 5. |  |  | q              ∧E-4
 6  |  |  p⊃q 
    |  +
 7  |  |_ ¬p∧¬q            H
 8  |  |  |_  p            H
 9  |  |  |  :
10  |  |  |  :            
11  |  |  |   _!_          
12  |  |  |   q            X-11 (or _!_E-11)
13  |  |  p ⊃ q            ⊃I 8-11
14  |  p ⊃ q               ∨E 1,2-6,7-12