# Help with discrete mathematics - inference and logical equivalence

I'm doing textbook homework for discrete mathematics and I'm struggling to understand how to solve the following practice problem. I would understand if they gave me variables and asked me to use premises to construct a specific argument, but I don't understand how to reduce it just to a value of false? Any help or hints would be appreciated.

Question:

Using only the rules of inference and the logical equivalences, show that the following argument is a contradiction by reducing it to a value of "False". You may assume that all the premises given are true.

𝑎 → 𝑏

¬𝑏 ∧ 𝑐

¬𝑎 → 𝑑

𝑑 → ¬𝑒

𝑒 ∧ f

• It seems like these are five premises, but you haven't given the conclusion. Is this really all there is to the exercise? It doesn't make sense to me. – saulspatz Feb 15 at 15:15
• From 3 and 4 derive $\lnot a \to \lnot e$. Then use it and 1 with $a \lor \lnot a$ to apply Dilemma to derive $\lnot e \lor b$. – Mauro ALLEGRANZA Feb 15 at 15:20
• Yeah, this is all that was given. I don't really understand it. – Eagerissac Feb 15 at 15:23
• Now you have $\lnot e \lor b$ with $\lnot b$ from 2 and $e$ from 5. – Mauro ALLEGRANZA Feb 15 at 15:23
• Is it enough to proceed, or do you need more hints ? – Mauro ALLEGRANZA Feb 15 at 15:51

The implication $$a \to b$$ is equivalent to $$\lnot a \lor b$$
Similarly, the other two implications can be written as $$a \lor d$$ and $$\lnot d \lor \lnot e$$
The two conjunctions $$\lnot b \land c$$ and $$e \land f$$ can only be true, if all four operands are true: $$\lnot b$$, $$c$$, $$e$$ and $$f$$
This simplifies $$\lnot a \lor b$$ to $$\lnot a$$
$$\lnot d \lor \lnot e$$ is simplified to $$\lnot d$$ which in turn simplifies $$a \lor d$$ to $$a$$.
The last step unveils a contradiction, as $$\lnot a$$ was found to be true earlier on.