TAOCP 7.1.1 Exercise 1
Let $x$ denote the proposition “it was so,” and let $y$ denote the proposition “it would be so” (equivalently, the consequent asserted under the condition that $x$ holds).
Exercise 1. [15] $[15]$ (Lewis Carroll.) Make sense of Tweedledee's comment, quoted near the beginning of this section. [Hint: See Table 1.]
Verified: yes
Solve time: 4m04s
Solution
Let $x$ denote the proposition “it was so,” and let $y$ denote the proposition “it would be so” (equivalently, the consequent asserted under the condition that $x$ holds). Tweedledee’s sentence is a paraphrase of a conditional statement of the form $x \supset y$.
In Table 1, the implication operator $\supset$ has truth table $1101$, ordered as $f(0,0),f(0,1),f(1,0),f(1,1)$. This means
$$ x \supset y = \begin{cases} 1 & \text{if } x=0,\ y & \text{if } x=1. \end{cases} $$
Equivalently, using the representation in Table 1,
$$ x \supset y = \bar{x} \vee y. $$
Tweedledee’s statement “if it was so, it might be; if it were so, it would be” asserts that when the antecedent $x$ is false, the conditional imposes no restriction on $y$. In Boolean terms, when $x=0$, the value of $x \supset y$ is $1$ for both $y=0$ and $y=1$, so both “might be” and “would be” are permitted.
The final clause “but as it isn’t, it ain’t” corresponds to the case $x=0$, where no contradiction arises regardless of the truth value of $y$. This matches the row $f(0,0)=1$ and $f(0,1)=1$ in the truth table of $\supset$.
When $x=1$, the implication reduces to $y$, since $f(1,0)=0$ and $f(1,1)=1$. Thus the conditional forces $y$ to match the asserted outcome whenever the antecedent holds.
Therefore Tweedledee’s reasoning restates that the material implication $x \supset y$ is defined by
$$ x \supset y = \bar{x} \vee y, $$
so it is true whenever $x=0$, and when $x=1$ its truth value is exactly that of $y$.
This completes the proof.
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