Show that given a contradiction you can assume anything

Question

In natural deduction there is a rule that says given a contradiction you can assume p i.e., $\frac{\bot}{p}$. I know that is a case when you show the soundness of the natural deduction system.

I tried to use induction to prove it. Here is my idea:

Suppose that $\forall j<i, \alpha_1, ...,\alpha_n \models \gamma_j $. A proof of natural deduction looks like this:

Step 1 $\gamma_1$ ... Step j $\gamma_j ...$ Step m $\gamma_m$...Step i $\gamma_i=\gamma_j\wedge\gamma_m$ by the introduction of $\wedge$, so $\alpha_1, ...,\alpha_n \models \gamma_j$ and $\neg\alpha_1, ...,\alpha_n \models \gamma_m$ by hypothesis. So by definition of $\wedge$ and $\models$, $(\alpha_1\wedge\neg\alpha_1), ...,\alpha_n \models \gamma_i$

$\therefore \bot\models\gamma_i$

Answer 1

We can use a truth table to prove that for any logical propositions $A$ and $B$ (they need not be related in any way), we have:

$A\land \neg A \implies B$.

Using a form of natural deduction, we have:

Answer 2

Short answer : the inference rule holds because, from a semantic point of view , the conditional

(contradictory antecedent) $\rightarrow$( whatever consequent )

is always a tautology.

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• Consider this reasoning.

Premise 1 : P&Q

Premise 2 : ~P

Conclusion : R.

Note that here, proposition R is chosen arbitrarily. I could have chosen any proposition instead of R, even ~(P&~P) , even ~Q, absolutely whatever. This is what I am going to show.

• In case the conjunction of the premises imply logically the conclusion, the reasoning is valid.

In other words, in case the conditional

((P&Q)&~P) $\rightarrow$ R

is a tautology ( a formula that is true in all possible situations/ interpretations), the reasoning is valid.

• By building the truth table of this formula, you will see that in all possible situations ( that is, in all the rows of the truth table) the antecedent - namely (P&Q)&~P - is false.

Now, the truth table of the " if ...then" operator tells you that : if the antecedent of a conditional is false ( lines 3-4 of the ->'s defining truth table) the whole conditional is automatically true.

In the particular case of our formula's truth table, every row corresponds either to line 3 or line 4 of the truth table defining the --> operator.

This is why the formula is true in all possible cases.

• Conclusion : the premises imply logically the conclusion, and the reasoning is valid, because

( .... contradictory antecedent) $\rightarrow$ ( whatever consequent )

is always a tautology.