P unless Q, in which case R

Question

I changed the variables in the title to make it easier in a google search in case anyone else needed to look up this problem.

In Leblanc and Wisdom's Deductive Logic, Third Edition, there is the following truth-table on page 230:

I can't seem to figure out what the formula for $A_1$ unless $A_2$, in which case $A_3$ would be. For instance,

(i) $(A_2 \vee A_1) \supset A_3$

does not fit the entry for the truth value assignment: $A_1$ as F, $A_2$ as F and $A_3$ as F

(ii) $ \neg(A_2 \supset A_3) \supset A_1$ does not fit the entry for the truth value assignment: $A_1$ as T, $A_2$ as T and $A_3$ as F

(iii) $A_1 \supset \neg(A_2 \supset A_3)$ does not fit the entry for any truth value assignment: $A_2$ as T and $A_3$ as F

(iv) $A_3 \supset (A_1 \vee A_2)$ does not fit the entry for the truth value assignment: $A_1$ as F, $A_2$ as F and $A_3$ as F

Does anyone know how to translate this statement or could it be that the book had a misprint in the last column of the truth-table?? Thanks in advance!!

Answer 1

Tip: 'unless' can be replaced with 'if not'.

So: '$A_1$ unless $A_2$, in which case $A_3$' becomes: '$A_1$ if not $A_2$, but if $A_2$ then $A_3$'.

'But' is of course 'and', and so we get:

$(\neg A_2 \rightarrow A_1)\land (A_2 \rightarrow A_3)$

Using your notation:

$(\neg A_2 \supset A_1) \land (A_2 \supset A_3)$

Notice that if we commute:

$(A_2 \supset A_3)\land (\neg A_2 \supset A_1)$

we get what the book identified as the equivalent '$A_3$ if $A_2$, otherwise $A_1$'

Answer 2

The formulae is rather simple guided by the comment $A_3$ if $A_2$, otherwise $A_1$. $$(A_3 \wedge A_2) \vee (A_1 \wedge \neg A_2) $$