demonstrate the following argument argument to be demonstrated:
$A ∨ (B ∧ C)$
$B → D$
$C → E$
$D ∧ E → A ∨ C$
$¬A$
$∴ C$
My attempt at this mathematical demonstration was as follows:
$A ∨ (B ∧ C)$
$≡ (A ∨ B) ∧ (A ∨ C)$
$≡ (A ∨ A) ∧ (B ∨ C)$
$≡ A ∧ (B ∨ C)$
$≡ (A ∧ B) ∨ (A ∧ C)$
 A: Note the first and last givens,  you are given "A or" something, and then "Not A"  That immediately gives you that the other part of the not must be true, so you get B and C.   Those then lead to give you D and E from the next two givens.   That then leads to A or C.  But once again, we know A is false, so...
A: In your attempt, the 3rd line is incorrect. If A is false then $A\lor (B\land C) $ (1st line) is equivalent to $B\land C.$ But if $A$ is false then $A\lor (B\lor C) $ (3rd line)  is equivalent to $B\lor C.$ Also, you must incorporate some of the given premises into the argument  to conclude that $C$ is true. Actually,  the first premise $A\lor (B\land C) $ and the fifth premise $\neg A$ are sufficient.
Rule: If $\neg X$ then $(X\lor Y)\implies Y.$
Rule: $\neg ((\neg A)\land A).$
Applying these when  $X$ is $((\neg A)\land A)$ and when  $Y$ is $(\neg A)\land (B\land C),$ we have $$[\,(\neg A)\land (A\lor (B\land C))\,]\implies [\,((\neg A)\land A))\,\lor\, ((\neg A)\land (B\land C))\,]\implies $$ $$\implies [\,(\neg A)\land (B\land C)\,]\implies$$ $$\implies  [\,(B\land C)\,]\implies C$$
A: This is a deductive reasoning question, rather than a propositional calculus one.
You are promised: $A\lor(B\land C)$, $B\to D$, $C\to E$, $(D\land E)\to (A\lor C)$, and $\lnot A$.
Well, because $A\lor (B\land C)$ and $\lnot A$, therefore $B\land C$, via Disjunctive Syllogism.
Because $B\land C$, therefore $B$ and $C$, via Simplification.
Because $B$ and $B\to D$ , therefore $D$, via Modus Ponens.
Bec...  and so on.
