I am trying to prove, using a series of logical equivalence rules, that the following formula is a tautology:
$$[a∧(a→b)∧(b→c)]→c$$
Yet despite numerous successes on other tautologies and logical equivalences proofs, I find myself getting stuck on this rather obvious syllogism.
A quick and easy truth table reveals this formula to be true, so I know that I am making a mistake somewhere.
Please note, I am not formally studying Propositional or Predicate Logic. Rather, I am studying this topic in my own time for a bit of fun.
We need the basic tautology of $p\to q\equiv\lnot p\lor q$, which only really needs truth tables to prove it.
\begin{array}{c|c|c|c|c} p&q&\lnot p&p\to q&\lnot p\lor q\\ \hline F&F&T&T&T\\ F&T&T&T&T\\ T&F&F&F&F\\ T&T&F&T&T \end{array}
$p\to q$ is always true except for when the premise $p$ is true and the implication $q$ is false, and this is exactly the same truth table as $\lnot p\lor q$.
Then we have $(a)\land (\lnot a\lor b)\land(\lnot b\lor c)$. All $3$ brackets need to be true for the LHS to be true, and for this to happen we therefore need $a\land\top=\top$, which in turn implies we need $b\land\top=\top$, and this in turn implies $c\land\top=\top$. With $c\land\top=\top$ we are done, in that if the LHS is false, the implication is true.
Another way of looking at it is that $a\land (\lnot a\lor b)\land(\lnot b\lor c)\equiv a\land b\land c\equiv c$.