# Proving $C \vdash D \lor \neg D$ using natural deduction, and WITHOUT any additional hypotheses/assumptions?

I proved $(A \land \neg A) \vdash B$ by doing the following:

\begin{array}{l l l} 1. & A \land \neg A & (\text{premise}) \\ 2. & A & (1, \text{ simplification}) \\ 3. & A \lor B & (2, \text{ addition}) \\ 4. & \neg A \land A & (1, \text{ commutative property}) \\ 5. & \neg A & (4, \text{ simplification}) \\ 6. & B & (3, 5, \text{ disjunctive syllogism}) \end{array} I heard it's also possible to prove $C \vdash D \lor \neg D$ like this - without using any additional hypotheses or assumptions. Could you guys give me some hints?

My natural deduction has:

• modus ponens
• modus tollens
• hypothetical syllogism
• disjunctive syllogism
• constructive dilemma
• simplification $((p \land q) \vdash p)$
• conjunction $(p, q \vdash p \land q)$
• addition $(p \vdash p \lor q)$
• absorption $(p \supset q \vdash p \supset (p \land q))$
• de Morgan's rule
• commutative property
• associative property
• distributive property
• double negation
• transposition
• material implication
• material equivalence
• exportation
• tautology ($p$ can be replaced with $p \land p$ and also the other way / $p$ can be replaced with $p \lor p$ and also the other way)
• Please, note that the rules for Natural Deduction are different. – Mauro ALLEGRANZA Jul 6 '17 at 7:31
• This is not natural deduction. I wish authors would quit calling every little system they invent that. – DanielV Jul 6 '17 at 9:57