# Task: Prove following with rules of inference (do not use truth table)

$$(A \vee B \rightarrow C) \wedge (C \rightarrow D \wedge C) \wedge (C \rightarrow D) \rightarrow (A \rightarrow D)$$

# Attempt to prove

1. $$A \vee B \rightarrow C$$ (premise)
2. $$C \rightarrow D \wedge C$$ (premise)
3. $$C \rightarrow D$$ (premise)
4. $$A \vee B \rightarrow D \wedge C$$ (Hypothetical syllogism 1,2)
5. $$\neg(A \vee B) \vee (D \wedge C)$$ (Logical equivalence 4)
6. $$(\neg A \wedge \neg B)\vee (D \wedge C)$$ (Logical equivalence 5)
7. $$\neg A \vee D$$ (Add 6)
8. $$A \rightarrow D$$ (Logical equivalence)

Q.E.D.

Is my proof valid?

• Either the formula or the proof are incorrect, you have an $\lor$ followed by an $\land$, but you interpreted them all as premises. – Luke Collins Jun 11 '19 at 14:36
• @LukeCollins There would be typo in the formula, I'll fix that – Tuki Jun 11 '19 at 14:37
• What do you get when you evaluate the formula with A being true and all the rest being false? – DanielV Jun 11 '19 at 14:51

I'm not sure how technical you want to be, but you actually gave the proof for $$A\lor B\to C, C\to D\land C, C\to D \vdash A\to D,$$ not what you showed. You would have to use $$\land$$-elimination to obtain what you dubbed the "premises" separately. This is the difference between a "proof" (in a metalanguage) and a "true implication".
Have you encountered the symbol $$\vdash$$ before? If not, don't worry about my point. As long as you only applied the rules you are allowed to apply, then the actual reasoning behind the proof you provided appears to be correct.
• Ah okay. The symbol $A\vdash B$ is basically like an inline version of $\frac AB$, so you actually do have to make some changes. You do not have three premises, but one (the whole thing before the $\to$). You need to first separate the conjoined statements using the Simp rule. – Luke Collins Jun 11 '19 at 14:56