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

In this truth table, we see that $(\neg P \rightarrow \bot)$ is equivalent to $P$. The line where $\neg P$ is $\text{true}$ is the last row and there $(\neg P \rightarrow \bot)$ is $\text{false}$.

In a proof by contradiction we assume that $\neg P$ is $\text{true}$ and show that this leads to a contradiction. Is this captured in the above truth table? How is the truth table connected to proof by contradiction?

The problem I have is that in a proof by contradiction we start by assuming $\neg P$ is $\text{true}$ and this leads to a contradiction. But I do not see how this is captured in the truth table?

EDIT: Alternatively we could l look at $P\rightarrow Q$ and we get the truth table:

$\begin{array}{|c c|c|c|c|} P & Q & (P \rightarrow Q)&(P\wedge \neg Q)&(P\wedge \neg 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} $

user394334
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1 Answers1

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

In this truth table, we see that $(\neg P \rightarrow \bot)$ is equivalent to P.

The structure of a proof by contradiction is $$(\neg P \rightarrow \bot) \;\text{ logically implying }\;P$$ (no biconditional or equivalence is involved in this skeleton), which is captured by the fact that every row of your truth table with a '$T$' in column 3 also has a '$T$' in the column 1. The contradiction in column 3 is derived when we mathematically obtain two statements such that one is a negation of the other.

in a proof by contradiction we start by assuming $\neg P$ is true and this leads to a contradiction. But I do not see how this is captured in the truth table?

The truth table captures only the logic of proof by contradiction; the meat of the proof (the details of deriving the contradiction) relies on some language and axioms of mathematics that a truth table, being a propositional-logic tool, does not attempt to capture; for example, $P$ might stand for $\forall x\;\big(\phi(x)\to\psi(x)\big).$

EDIT corresponding to the OP's edit

$P\rightarrow Q$

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

This truth table is merely a special case of the above: here, $(P{\implies} Q)$ is being proven by contradiction, by supposing $P$ and assuming $\lnot Q;$ its structure is captured by the fact that every row with a '$T$' in the 5th column also has a '$T$' in the 3rd column.

However, this special case does not fully represent the mathematics proof, unlike the above general case where $P$ can symbolise any mathematical statement. Here, $$P\to Q$$ symbolises merely the predicate (for example) $$\phi(x)\to\psi(x),$$ so there's an additional step (Universal Introduction, usually tacit, but where you might explicitly note that you are generalising the result for a representative/arbitrary variable $a$ to every variable $x$) to finally derive the intended mathematical statement (for example) $$\forall x\;\big(\phi(x)\to\psi(x)\big).$$

ryang
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