Basic Tautology Question I'm reviewing an old exam to study for my upcoming final, and one of the questions is this:
"Show that $a∨b \rightarrow¬a \rightarrow b$ is a tautology"
My professor gave us this definition for tautology: "A propositional formula φ is a tautology if for all valuations ν, ν(φ)=⊤"
So... I don't understand what the statement $a∨b \rightarrow¬a \rightarrow b$ means. It certainly can't mean that if $a∨b$ is true, then $¬a$ is true, because that seems indeterminable. Is the statement saying that if $a∨b$ is true, AND $¬a$ is true, then $b$ must be true? The arrows just feel sort of ambiguous to me, because I'm tempted to read them as "implies", but that doesn't seem correct.
 A: I think the way to go with this one is to fiddle with the algebra of it to show that it ultimately is a statement that is always true.
Generally, we consider $a\to b\to c$ to be $a\to (b\to c)$. Assuming this we have:
$$
a∨b→¬a→b \equiv a∨b→(¬a→b) \equiv a∨b→(\lnot\lnot a\lor b) \equiv \lnot(a∨b)\lor(a\lor b)
$$
And the last is evidently a tautology.
A: If I understand the question, this is primarily a question about syntax.  The formula
$$a \vee b \rightarrow ¬a \rightarrow b$$
is shorthand for
$$a \vee b \rightarrow (¬a \rightarrow b)$$
since by convention, $\rightarrow$ is taken to be right associative, e.g. $\alpha \rightarrow \beta \rightarrow \gamma \rightarrow \delta \rightarrow \epsilon$ stands for $\alpha \rightarrow (\beta \rightarrow (\gamma \rightarrow (\delta \rightarrow \epsilon)))$.  Then the question becomes intelligible.
Note that the above formula is equivalent to
$$((a \vee b) \land ¬a) \rightarrow b$$
under the reversible, derivable rule
$$ \chi \rightarrow (\phi \rightarrow \psi) \over (\chi \land \phi) \rightarrow \psi $$
which makes it easy to see why the formula is true, and a bit easier to check the truth table.
A: Like you said, a tautology in Sentence Logic, where the internal structure of sentences does not matter, it is a sentence that cannot be falsified by any assignment of truth values to the component sentences. This is maybe a weakness , because of the simplicity of Sentence calculus. If you were doing predicate logic, a.k.a, first-order logic, then a tautology takes on a different meaning where the actual content of the component sentences affects the truth value of the overall sentences. For this case, you just want to show that the total output of a truth table with any of the $4$ assignments to $a,b$ will produce a true sentence.
A: a or b implies ( not a implies b)
so if a or b is true and a is not true, b must be true
A: Assuming that we have disambiguated the formula with the correct insertion of parentheses, the easiest way to check for "tautologuesness" is to use truth-tables :

$\nu (p \lor q) = F$ iff $\nu(p) = \nu(q)=F$
$\nu (p \rightarrow q) = F$ iff $\nu(p)=T$ and $\nu(q)=F$

and, of course :

$\nu(\lnot p) = F$ iff $\nu(p) = T$.

A: IF we're right and this statement is actually $(a \vee b) \rightarrow (\sim a \rightarrow b)$, do a truth value analysis on it. Case I: $a:T$. Then the statement collapses to $T \to T$ (why?), which is true.
Case II: $a:\perp$. Then we get $b \rightarrow (T \rightarrow b)$, which is $b \rightarrow b$ (why?), which is also always true. (One last time: why?)
