Q:

Tautological Implications

5. Prove that if for some formulae $A$ and $B$ it is the case that $\top\vDash_{\text{taut}}A\to B$ and $B\vDash_{\text{taut}}\bot,$ then it is also the case that $\vDash_{\text{taut}}\neg A$.

Here, using truth tables or truth-table tricks, you will show that if you have the left side of the “then” in the statement, then you must have the right side as well.

6. Prove that, for any formula $A$, we have $\top\vDash_{\text{taut}}A$ iff $\vDash_{\text{taut}}A$.

Requirement: Clearly state the assumption and the required conclusion in each direction of the “iff”, and then proceed to establish the conclusions in each direction.

Original image.
For question 5, It's telling me to use truth tables, so it's asking me to show that the 2 tautological implications left of the then are valid, then check the right? Maybe I'm going about it wrong.
Same problem with question 6, no idea where to go.
Any help would be appreciated, I finished the rest of the assignment last 2 I'm stuck on.
Thanks

A:

HINT: I’m assuming that your notation has been defined so that $P\vDash_{\text{taut}}Q$ means something like this: 

If we make a joint truth table for $P$ and $Q$, then $Q$ is true in every line in which $P$ is true. That is, every assignment of truth values to the propositional variables in $P$ and $Q$ that makes $P$ true also makes $Q$ true.

Similarly, $\vDash_{\text{taut}}Q$ means:

$Q$ is true in every row of a truth table for $Q$. That is, $Q$ is true for every assignment of truth values to its propositional variables.

(If your definitions are substantially different, a different answer will be needed.) 
Then (6) is easily argued from the fact that $\top$ is true for every possible assignment of truth values. I’d prove it first and then go back to (5), where I’d first use (6) to replace $\top\vDash_{\text{taut}}A\to B$ by the simpler (but now known to be equivalent) $\vDash_{\text{taut}}A\to B$. Then imagine making a joint truth table for $A,B,A\to B,\bot$, and $\neg A$. Explain what $\vDash_{\text{taut}}A\to B$ tells you about the truth table: in each row either $\text{something}$ or $\text{something else}$. Similarly, see what $B\vDash_{\text{taut}}\bot$ tells you about certain rows of the truth table. Then put that information together to deduce that $\neg A$ must be false in every row.

