[Part 4] Raymond Smullyan's Certified Knights and Knaves puzzles Puzzle #4 - Going backwards again

For our fourth puzzle, we will solve Problem 4 on pages 41-42 of the 2013 Dover edition [ISBN: 978-0486497051] of Raymond Smullyan's 2013 book entitled "The Gödelian Puzzle Book - Puzzles, Paradoxes & Proofs". 

This puzzle reads:

I once came across a native who made a statement such that I could deduce that he must be certified (either a certified knight or a certified knave), but there was no possible way to tell whether he was a knight or a knave. 

What statement would work?

Just like in the previous post, we build a truth table in which the last column contains the value True in the only two rows that correspond to the scenarios allowed by the puzzle statement:

\(A\)	\(Acert\)	\(\varphi\) = ?	\(A \leftrightarrow \varphi\)
\(F\)	\(F\)		\(F\)
\(F\)	\(T\)		\(T\)
\(T\)	\(F\)		\(F\)
\(T\)	\(T\)		\(T\)

Solving the puzzle now means finding a formula \(\varphi\) that correctly completes this table.

Since the main connective in \(A \leftrightarrow \varphi\) is a biconditional, \(\varphi\) must have the same truth value as \(A\) in the second and fourth rows (where the biconditional is true) and a different truth value from \(A\) in the first and third rows (where the biconditional is false).

Therefore, the almost-full table must look like:

\(A\)	\(Acert\)	\(\varphi\) = ?	\(A \leftrightarrow \varphi\)
\(F\)	\(F\)	\(T\)	\(F\)
\(F\)	\(T\)	\(F\)	\(T\)
\(T\)	\(F\)	\(F\)	\(F\)
\(T\)	\(T\)	\(T\)	\(T\)

Finally, the table will be complete (and the puzzle solved!) once we have found a formula \(\varphi\) that works:

\(A\)	\(Acert\)	\(\varphi\) = ?	\(A \leftrightarrow \varphi\)
\(F\)	\(F\)	\(T\)	\(F\)
\(F\)	\(T\)	\(F\)	\(T\)
\(T\)	\(F\)	\(F\)	\(F\)
\(T\)	\(T\)	\(T\)	\(T\)

In the most general case, we can assume that \(\varphi\) will depend on the two propositions in this puzzle, namely \(A\) and \(Acert\).
By inspection of the table above, we see that \(\varphi\) is true exactly when \(A\) and \(Acert\) have the same truth value.
Therefore, \(\varphi\) can simply be chosen to be:
\(A \leftrightarrow Acert\)
which could have come from the English statement: 

 "I am a knight if and only if I am certified."

And this statement is indeed logically equivalent to the answer given by Smullyan on page 44 of his book:

"I am either a certified knight or an uncertified knave."