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How is proof by contradiction connected to propositional calculus?

When using proof by contradiction to prove $P,$ we assume that $\neg P$ is $\text{true}$ and show that this leads to a contradiction. How does the following truth table capture this? I see that it shows us that $(\neg P \rightarrow \bot)$ is equivalent to $P;$ I also see that the row where $\neg P$ is $\text{true}$ is the last one, and that in that row $(\neg P \rightarrow \bot)$ is $\text{false}.$

\begin{array}{|c |c|c|} P & \lnot P& (\neg P \rightarrow \bot)\\ % Use & to separate the columns \hline % Put a horizontal line between the table header and the rest. \text{T} & \text{F} & \text{T}\\ \text{F} & \text{T} & \text{F}\\ \end{array}

Alternatively, how does the following truth table capture proving $P\rightarrow Q$ using contradiction ? \begin{array}{|c c|c|c|c|} P & Q & (P \rightarrow Q)&\lnot(P\to Q)&\lnot(P\to Q)\rightarrow \bot\\ % Use & to separate the columns \hline % Put a horizontal line between the table header and the rest. \text{T} & \text{T} &\text{T}& \text{F} &\text{T}\\ \text{T} & \text{F} & \text{F}& \text{T} &\text{F}\\ \text{F} & \text{T} &\text{T}& \text{F} &\text{T}\\ \text{F} & \text{F} &\text{T}& \text{F} &\text{T}\\ \end{array}

ryang
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user394334
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1 Answers1

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How is proof by contradiction connected to propositional calculus?

The truth table \begin{array}{|c|c|} P & (\neg P \rightarrow \bot)\to P\\ % Use & to separate the columns \hline % Put a horizontal line between the table header and the rest. \text{T} & \text{T} \\ \text{F} & \text{T} \\ \end{array} being filled in its main column with only $T$'s shows that $$(\neg P \rightarrow \bot) \;\text{ logically implies }P.$$ This tautology is the implicit premise—and backbone—of proof by contradiction, and allows us to conclude from our derivation $(\neg P \rightarrow \bot),$ i.e.,

we assume that $\neg P$ is $\text{true}$ and show that this leads to a contradiction,

that $P$ must in fact be true.

How do the following truth tables capture this? \begin{array}{|c |c|c|} P & \lnot P& (\neg P \rightarrow \bot)\\ % Use & to separate the columns \hline % Put a horizontal line between the table header and the rest. \text{T} & \text{F} & \text{T}\\ \text{F} & \text{T} & \text{F}\\ \end{array}

Every row of this truth table that has a '$T$' in the last column also has a '$T$' under the statement that we're trying to prove (i.e., in column 1).

I see that the row where $\neg P$ is $\text{true}$ is the last one, and that in that row $(\neg P \rightarrow \bot)$ is $\text{false}.$

You're saying that every row where $P$ is false has $(\neg P \rightarrow \bot)$ also false. This is equivalent to my previous paragraph.

I see that it shows us that $(\neg P \rightarrow \bot)$ is equivalent to $P$

Sure, but the $$(\neg P \to \bot)\Longleftarrow P$$ part isn't required for proof by contradiction.

\begin{array}{|c c|c|c|c|} P & Q & (P \rightarrow Q)&\lnot(P\to Q)&\lnot(P\to Q)\rightarrow \bot\\ % Use & to separate the columns \hline % Put a horizontal line between the table header and the rest. \text{T} & \text{T} &\text{T}& \text{F} &\text{T}\\ \text{T} & \text{F} & \text{F}& \text{T} &\text{F}\\ \text{F} & \text{T} &\text{T}& \text{F} &\text{T}\\ \text{F} & \text{F} &\text{T}& \text{F} &\text{T}\\ \end{array}

Similarly as for the previous table: every row of this truth table that has a '$T$' in the last column also has a '$T$' under the statement that we're trying to prove (i.e., in column 3).

ryang
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